Group theory 1

Groups

[https://crypto.stanford.edu/pbc/notes/group/group.html].

We define a group $G^{\ast}$ as a set $G$ and a binary operation $\cdot$ having the following properties:

1. For all $x, y \in G, x \cdot y \in G$. This is normally referred to as 'algebraic closure'.
2. There exist an identity element $1 \in G$ such that $x \cdot 1 = x$ for all $x \in G$.
3. The operation defined $(\cdot)$ is associative, that is, for all $x, y, z \in G$ we have $x(yz) = (xy)z$.
4. For all $x \in G$ there exists an element $x^{-1}$ with $xx^{-1} = x^{-1}x = 1$ (the inverse). I proved this property here [https://github.com/Z323323/From-Fermat-to-the-group-theory?tab=readme-ov-file#proof-of-existence-and-uniqueness-of-inverses-in-z_p-and-z_phinast]. Note also that the formal definition of group doesn't imply we have a finite group and it's therefore more general. The linked proof is about finite multiplicative groups, which are the main interest for computer scientists.

• If only $1,3$ holds for $G^{*}$ then $G^{\ast}$ is a semi-group.
• If only $1, 2, 3$ holds for $G^{*}$ then $G^{\ast}$ is a monoid.

In general if you see $G^{\ast}$ around in papers this will refer to a multiplicative group.

If $G^{\ast}$ is commutative then

5. For all $x, y \in G, xy = yx$

and we call it abelian.

Axioms

1. Closure (not really trivial if you played around with Zn.py using $n$ non-prime).
2. Associativity (trivial).
3. Right and left cancellation:
   • $ax = bx$ iff $a = b$. This one is not that trivial if you think about finite groups and therefore modular arithmetic [https://github.com/Z323323/From-Fermat-to-the-group-theory?tab=readme-ov-file#cancellation-law].

Order

The order of an element $g$ in a group $G$ is the smallest positive integer $k$ such that $g^{k} = 1$. We know this exists for all elements of $Z_{\phi(n)}^{\ast}$ (finite groups), while it doesn't for all elements of infinite groups (holds only for $1$, at my current perspective).

Subsets

A subset $H$ of $G$ which satisfy the previous properties is called a subgroup of $G^{*}$. Every finite group has two improper subgroups which are defined by the subsets $\{1\}$, and $G$. All other subgroups are defined as proper subgroups.

We can write $H \leq G$ to refer to any subgroup of $G^{*}$ and if $H \neq G$ then $H < G$.

Homomorphism and isomorphism

An homomorphism is a map $f : G → H$ between two different groups where

$f(x)f(y) = f(xy)$

for all $x, y \in G$. If $f$ is bijective then $f$ is called isomorphism.

Let's clarify this concept, since it looks really abstract, and it is indeed, and it's also really important in math.

First thing to understand is the meaning of map, and the different kinds of map in mathematics.

1. map = function. A function is a map which associates elements from a set (domain) to some elements of another set (codomain). The actual elements mapped will be elements $\in$ codomain, which are called image of the function/map, and so we could have image = codomain or not.

2. A map/function has $2$ foundation constraints which defines it.

   • It can't have two mappings starting from the domain, it's exactly one map for every element of the domain, while it can for example has two elements which map the same element of the codomain/image.
   • Every element of the domain must be mapped (it's not true for the codomain).

3. A map/function has different properties which define it's structure, it can be

   • surjective: every element of the codomain is mapped.
   • injective: every element of the domain maps $1$ and only $1$ element of the codomain.
   • bijective: if it's both surjective and injective.

Now let's get back to our omomorphism definition. An omomorphism is a particular mapping/function where its definition holds, that is, where the mapping $G → H$ must satisfy

$(x \in G → H)(y \in G → H) = (xy \in G → H)$
$->$
$(x \in G → H)?_2(y \in G → H) = (x?_1y \in G → H)$
$->$
$(a \in H)?_2(b \in H) = (x?_1y \in G → H)$

Here lie a couple behaviours which could not be straightforward at first sight, we can see that there are two implied relations $?_{1, 2}$ between $(a)?_2(b), a, b \in H$ and $x?_1y, x, y \in G$. This relation strongly affects the resulting property, and the reason of subscripts is because of the implied definitions

$G^{?_1} = (\{x, y, \dots \}, ?_1)$
$H^{?_2} = (\{a, b, \dots \}, ?_2)$

and because that relation (binary operation) will be defined individually for elements which belong to $G$ or $H$. Now, if we consider $G$ and $H$ as multiplicative finite groups then the implied relation is exactly the same because the binary operation defined for multiplicative finite groups is the same, but this is not true in general.

Example

Imagine the infinite group $G$ defined over the positive integers towards infinity and $H$ being a finite group $Z_{n}^{\ast}$ of order $n - 1$, thus $n$ is prime, we have

$G = (\{1, 2, \dots, \infty \}, ?_1 = \cdot)$
$H = (\{1, 2, \dots, n - 1 \}, ?_2 = \cdot \mod n)$

We define the omomorphism $f : G → H$ where $f(x) = x \mod n$, then we have

$(x \in G → H)(y \in G → H) = (xy \in G → H)$
$->$
$(x \in G → H)?_2(y \in G → H) = (x?_1y \in G → H)$
$->$
$(x \mod n)?_2(y \mod n) = (x \cdot y \mod n)$
$->$
$(x \mod n) \cdot (y \mod n) \mod n = (x \cdot y \mod n)$

which holds since [https://github.com/Z323323/From-Fermat-to-the-group-theory?tab=readme-ov-file#multiplication-property].

This means that the multiplicative group of positive integers is omomorphic to the multiplicative group of positive integers $\mod n$ where $n$ is prime. Or stated in a different manner, there exists an omomorphism between the multiplicative group of positive integers and the the multiplicative group of positive integers $\mod n$ where $n$ is prime.

Now, since our focus will be directed towards finite multiplicative groups let's analyze the definition of homomorphism for two general finite groups $N = Z_{n}^{\ast}$, $M = Z_{m}^{\ast}$ where $n, m$ are prime numbers.

$N = (\{x, y, \dots, n - 1 \}, \cdot \mod n)$
$M = (\{a, b, \dots, m - 1 \}, \cdot \mod m)$

We can see that for multiplicative groups we can avoid the construction regarding the binary operator and just consider the modulo, then we define the omomorphism $f : N → M$, where $f(x)$ maps elements from the group $Z_{n}^{\ast}$ to $Z_{m}^{\ast}$.

$(x \in N → M)(y \in N → M) = (xy \in N → M)$
$->$
$(x \mod n → M)(y \mod n → M) = (xy \mod n → M)$
$->$
$((x \mod n) \mod m)((y \mod n) \mod m) \mod m = (xy \mod n) \mod m$

Note that when reasoning about multiplicative subgroups the previous form simplifies because the modulo is shared, and for example $((x \mod n) \mod n) = x \mod n$.

Some basic theorems collapsed

$(1)$

If $g \in G$ has order $o$, then $x^{m} = 1$ iff $o | m$. This is a direct consequence of [https://github.com/Z323323/Group-theory-elements?tab=readme-ov-file#proof-of-cyclicness-of-subgroups-and-uniqueness-of-each-element].

$(2)$

If $x \in G$ has order $mn$, where $m, n$ are coprime, then $x$ can be uniquely expressed in the form $x = uv$ where $u$ has order $m$ and $v$ has order $n$. Let $u = x^{bn}$ and $v = x^{am}$ such that $bn + am = 1$, by Bezout's Identity that equation always has a solution. This means that for example, if the solution of the identity is $bn - am = 1$ then

$\displaystyle x = \frac{x^{bn}}{x^{am}}$

This theorem could look very strange at first, you should note that we are making an abstraction considering $x$, because $x = g^{k}$ for some $k$ and

$bn - am = 1$
$->$
$\displaystyle g^{k} = \frac{g^{kbn}}{g^{kam}} = g^{k(bn - am)} = g^{k(1)} = g^{k} = x$

We can easily replace $bn - am = 1$ using the general form $bn + am = 1$ being aware we will either end up having $bn - am = 1$ or $am - bn = 1$ which produces the same outcome.

$(3)$

A non-empty subset $H$ of $G$ is a subgroup of $G$ iff $H$ is closed under multiplication (1). This is a direct consequence of the definition of subgroups I wrote at [https://github.com/Z323323/Group-theory-elements/blob/main/README.md].

$(4)$

A non-empty subset $H \subset G$ is a subgroup iff $H^{2} \subset H$. To get this imagine $H$ not being a subgroup of $G$. We have that $e^{2} \not\in H$ for some element $e \in H$ and therefore $H$ not being a subgroup of $G$.

$(5)$

For a subgroup $H$, for all $h \in H$ it must be that $hH = H = Hh$. Following [https://github.com/Z323323/Group-theory-elements?tab=readme-ov-file#multiplicative-groups-cyclic-subgroups-and-generators] $h$ must be some power of $e$ where $e$ is a generator for the subgroup; the lemma follows.

$(6)$

For any set $S \subset H$, $SH = H = HS$. This is exactly the same logic of the previous one.

$(7)$

A non-empty subset $H \subset G$ is a subgroup iff $H^{2} = H$. Follows from (4). You could also imagine $\langle h \rangle = H$, then shifting every element right by $1$ position clearly produces $H$.

$(8)$

Let $g \in G$ and $H \leq G$, then $g^{- 1}Hg$ is a subgroup of $G$ isomorphic to $H$. To figure this out quickly, remember that $G$ and $H$ are defined over the same modulo, this means that we are basically removing elements from $H$ and then reinserting them. Calling $R$ the resulting set, we will have $|R| = |H|$ and $R = H$.

Setting up a convenient formalization on subgroups

Analyzing $(2)$ I noticed we can make a convenient formalization.

Let $S$ be a subgroup of $G$, we can refer to every element of $S$ letting $g \in G$ generator for $G$ and writing

$g^{kx}$

where $x \geq 1$ and $k$ is a constant such that the order $o$ of $S$ is $\phi(n) / k$ for some $k | \phi(n)$ ($S$ is the subgroup defined by the roots of unity of $o$).

Example 1

Let $S$ be a subgroup of $G$ of order $o = \phi(n) / 2$, and $g$ a generator for $G$, then we can represent every element of $S$ using the general form

$g^{2x}$

Example 2

Let $G$ of order $\phi(n)$ where $\phi(n) = 2 \cdot 3 \cdot 13$; let $g$ a generator for $G$ and $S$ a subgroup of order $o = 13$, then we have

$\displaystyle o_S = \frac{\phi(n)}{2 \cdot 3}$

and we can define the general form for every element of $S$ as

$g^{6x}$

Lagrange's theorem

[https://crypto.stanford.edu/pbc/notes/group/lagrange.html].

Lemma

Let $H$ be a subgroup of $G$ and $r, s \in G$, then $Hr = Hs$ iff $rs^{- 1} \in H$, otherwise $Hr$ and $Hs$ have no element in common.

Proof

Let $r = g^{k_1x}, s = g^{k_2x}$, and the general form for any element of $S$ as $g^{kx}$. In order to have $Hr = Hs$ it must be that

$(k_1 + k)x = (k_2 + k)x$
$->$
$k_1 = k_2$

Under this result we can easily see that $rs^{- 1}$ translates to

$\displaystyle \frac{g^{k_1x}}{g^{k_2x}} = g^{(k_1 - k_2)x}$

and under $Hr = Hs$ we have $k_1 = k_2$ and

$g^{(k_1 - k_2)x} = 1 \in H$

which is always true since $1$ belongs to every subgroup.

Theorem

If $H$ is a subgroup of $G$ then $|G| = n|H|$ for some positive integer $n$ which is called index of $H$ in $G$. Furthermore, there exist $r_1, \dots, r_n \in G$ such that $G = Hr_1 \cup \dots \cup Hr_n$, and an equivalent construction holds for $H$ if the order of $H$ is a composite number.

Proof

The first statement derives directly from the general form for subgroups elements I wrote previously and

$\displaystyle index_H = \frac{o_G}{o_H}$

The second statement holds because, recalling the general form for the elements of $H$ subgroup of $G$ being $g^{kx}$, and $(2)$, we can use two elements which belong to two different subgroups having coprimes orders to build every element of $G$ such that the two orders multiplied equal the order of $G$. Under this last assumption $(o_G = o_Ho_S)$ lies the proof for this theorem, because setting a new hypothesis where all indexes of the subgroups multiplied equal $\phi(n)$ produces

$G = g^{xk_Hk_Sk_R\dots} = g^{x\phi(n)}$

which is the general form for the elements of $G$. This holds as long as there exist a generator for $G$. In case this doesn't exist it should be possible to generate some $Z_{\phi(n)}^{\ast}$ using an independent generator and a generator of order $2^{k - 2}$ where $\phi(n) = 2^{k - 1}$. We can expand this further considering other cofactors but for the moment I'm not doing this.

Corollary 1

Let $|G|$ be the order of $G$, and $g \in G$, then $g \in H$ and $|H| | |G|$. This is a direct consequence of the Euler's theorem and the general forms structure.

Corollary 2

Let $G$ a group of prime order, then $G$ has no subgroups while being cyclic by Euler's theorem (and corollary but the more I study these topics the more I realize Euler's theorem is enough to state cyclicality).

Additive (cyclic) groups

It is worth to spend a couple words on additive finite groups now. To define them we can just take the definition of multiplicative subgroups and change the identity element to $0$. It derives that the inverse of a number in an additive group is its negative. Also, it is quite easy to figure out that for any prime $n$, the order of any subgroup of $Z_{n}^{+}$ will be $n$, because, for example, imagine $Z_{17}^{+}$, it's quite simple to figure out that we will encounter $0$ every time we add any generator $17$ times and not before since no elements shares the $17$ cofactor before being multiplied by it. Let's now talk about uniqueness of elements and cases having $n$ which is not prime.

Example

Defining $Z_{6}^{+}$ and using $1, 2, 3, 4, 5$ as subgroups generators we get

1. $\{1, 2, 3, 4, 5, 0, \dots \}$
2. $\{2, 4, 0, 2, 4, 0, \dots \}$
3. $\{3, 0, 3, 0, 3, 0, \dots \}$
4. $\{4, 2, 0, 4, 2, 0, \dots \}$
5. $\{5, 4, 3, 2, 1, 0, \dots \}$

and we can derive a couple of simple properties. We have $1, 5$ being generators for the set and it's quite intuitive to derive that any generator will need to be coprime with the modulo. Now let's talk about uniqueness and orders. Elements which share the same cofactor(s) with the modulo will have the same order, and every element before $0$ will be different ensuring uniqueness of elements because, let $n = abc$, moving the problem to multiplicative groups we get

• $a \cdot bc \equiv 0 \mod abc$
• $ab \cdot c \equiv 0 \mod abc$
• $d \cdot abc \equiv 0 \mod abc$
• $aa \cdot bc \equiv 0 \mod abc$
• $ad \cdot bc \equiv 0 \mod abc$
• $abd \cdot c \equiv 0 \mod abc$
• $de \cdot abc \equiv 0 \mod abc$

where numbers on the left side of $\cdot$ represent subgroups generators for $Z_{abc}^{+}$, and numbers on the right represent their subgroup order (which prove everything said but uniqueness). It results pretty clear that uniqueness follows since we'll get different results until the $o + 1$ addition, similarly to multiplicative groups.

We can now state a couple rules.

Theorem 1

Every subgroup of a cyclic group is cyclic, and this works both for additive and multiplicative finite groups. This doesn't need a proof, since everything said.

Theorem 2

Every group which has composite order have proper subgroups. This doesn't need a proof either.

Isomorphisms between multiplicative and additive groups

Note that since the course analyzed is a general course on group theory a lot of things could become quite strange just because the definition of group, as we saw earlier, admits various interpretations. Since as I already mentioned we are mainly interested in finite groups, theorems will normally refer to finite multiplicative groups and finite additive groups which we are going to combine together from now on.

Lemma

If each element $1 \neq g \in G$ is of order $2$ then $G$ is abelian and isomorphic to $Z_{2}^{+} \times \dots \times Z_{2}^{+}$ and $|G|$ is a power of $2$.

Proof

Considering the scenario of the lemma proposed, the isomorphism for $|G| = 2^{2}$ can be considered as a mapping $Z_{\phi(8)}^{\ast} ↔ Z_{2}^{+} \times Z_{2}^{+}$, because

$Z_{\phi(8)}^{\ast} = (\{ 1, 3, 5, 7 \}, \cdot \mod 4)$
$->$
$|Z_{\phi(8)}^{\ast}| = |\{ 1, 3, 5, 7 \}| = 4$

and each element of $Z_{\phi(8)}^{\ast} \neq 1$ has order $2$ (see the link below to get why). Let's now analyse $Z_{2}^{+} \times Z_{2}^{+}$.

$Z_{2}^{+} \times Z_{2}^{+} = (\{ (0, 0), (1, 0), (0, 1), (1, 1) \}, + \mod 2)$
$->$
$|Z_{2}^{+} \times Z_{2}^{+}| = |\{ (0, 0), (1, 0), (0, 1), (1, 1) \}| = 4$

Since [https://github.com/Z323323/Group-theory-elements?tab=readme-ov-file#theorem] we know that $Z_{\phi(8)}^{\ast}$ has $1$ element $(1)$ of order $1$, and $3$ of order $2$. We can see that this is the same for $Z_{2}^{+} \times Z_{2}^{+}$ because, setting $(0, 0)$ as the identity we have

$Z_{2}^{+} \times Z_{2}^{+} = (\{ (0, 0), (1, 0), (0, 1), (1, 1) \}, + \mod 2)$
$->$
$\{ (0, 0), (1, 0) + (1, 0) = (0, 0), (0, 1) + (0, 1) = (0, 0), (1, 1) + (1, 1) = (0, 0) \}$

that is, every element has order $2$ except for $(0, 0)$ which has order $1$ and $1 ↔ (0, 0)$. Formalyzing produces

$(x \in Z_{\phi(8)}^{\ast} ↔ Z_{2}^{+} \times Z_{2}^{+}) + (y \in Z_{\phi(8)}^{\ast} ↔ Z_{2}^{+} \times Z_{2}^{+}) \mod 2 = (xy \in Z_{\phi(8)}^{\ast} ↔ Z_{2}^{+} \times Z_{2}^{+})$

which is quite unbelievable at first, that is, since $Z_{\phi(8)}^{\ast}$ doesn't have generators, it's true that every subgroup generator multiplied by another subgroup generator won't produce $1$. At the same time it must be that another element $e \in Z_{\phi(8)}^{\ast}$ will be produced, thus, for this particular example, another generator for some subgroup will be produced, this is reflected by $Z_{2}^{+} \times Z_{2}^{+}$ because

• $(0, 0) + X = X$
• $(1, 0) + (0, 1) = (1, 1)$
• $(1, 0) + (1, 1) = (0, 1)$
• $(0, 1) + (1, 0) = (1, 1)$
• $(0, 1) + (1, 1) = (1, 0)$
• $(1, 1) + (1, 0) = (0, 1)$
• $(1, 1) + (0, 1) = (1, 0)$

and

$(1, 0) + (0, 1) + (1, 1) = (0, 0)$

and

$3 \cdot 5 \cdot 7 \mod 8 = 1$

unbelievable right? And it's not over because since the maps

• $1 ↔ (0, 0)$
• $3 ↔ (1, 0)$
• $5 ↔ (0, 1)$
• $7 ↔ (1, 1)$

we have

• $(0, 0) + X = X$
• $(1, 0) + (0, 1) = (1, 1) ↔ 3 \cdot 5 \mod 8 = 7$
• $(1, 0) + (1, 1) = (0, 1) ↔ 3 \cdot 7 \mod 8 = 5$
• $(0, 1) + (1, 0) = (1, 1) ↔ 5 \cdot 3 \mod 8 = 7$
• $(0, 1) + (1, 1) = (1, 0) ↔ 5 \cdot 7 \mod 8 = 3$
• $(1, 1) + (1, 0) = (0, 1) ↔ 7 \cdot 3 \mod 8 = 5$
• $(1, 1) + (0, 1) = (1, 0) ↔ 7 \cdot 5 \mod 8 = 3$

Let's now analyse $Z_{\phi(16)}^{\ast} ↔ Z_{2}^{+} \times Z_{2}^{+} \times Z_{2}^{+} ↔ Z_{4}^{+} \times Z_{2}^{+}$, we have

$Z_{\phi(16)}^{\ast} = (\{ 1, 3, 5, 7, 9, 11, 13, 15 \}, \cdot \mod 16)$
$->$
$|Z_{\phi(16)}^{\ast}| = 8$

This case introduces some complexity, because since the $max - order$ $m_o = 4$ we'll have some generators which produces up to $2$ other generators before $1$. A quick check reveals $3$ produces $9, 11$ and $7$ has order $2$, and this means that we can just use $3$ and $7$ to get every other generator. For ex. setting the identity $(0, 0)$ and defining

• $3 ↔ (0, 1)$
• $7 ↔ (2, 2)$
• $->$
• $3^2 \mod 16 = 9 ↔ (0, 1) + (0, 1) = (0, 2)$
• $3^3 \mod 16 = 11 ↔ (0, 2) + (0, 1) = (0, 3)$
• $3^4 \mod 16 = 1 ↔ (0, 3) + (0, 1) = (0, 0)$
• $and$
• $7^{2} \mod 16 = 1 ↔ (2, 2) + (2, 2) = (0, 0, 0)$
• $then$
• $3 \cdot 7 \mod 16 = 5 ↔ (0, 1) + (2, 2) = (2, 3)$
• $3^{2} \cdot 7 \mod 16 = 15 ↔ (0, 2) + (2, 2) = (2, 0)$
• $3^{3} \cdot 7 \mod 16 = 13 ↔ (0, 3) + (2, 2) = (2, 1)$
• $also$
• $5^{2} \mod 16 = 9 ↔ (2, 3) + (2, 3) = (0, 2)$
• $5^{3} \mod 16 = 13 ↔ (0, 2) + (2, 3) = (2, 1)$
• $9^{2} \mod 16 = 1 ↔ (0, 2) + (0, 2) = (0, 0)$
• $13^{2} \mod 16 = 9 ↔ (2, 1) + (2, 1) = (0, 2)$
• $13^{3} \mod 16 = 5 ↔ (0, 2) + (2, 1) = (2, 3)$
• $13^{4} \mod 16 = 1 ↔ (2, 3) + (2, 1) = (0, 0)$
• $15^{2} \mod 16 = 1 ↔ (2, 0) + (2, 0) = (0, 0)$

$Z_{2}^{+} \times Z_{4}^{+} = (\{ (0, 0), (0, 1), (2, 3), (2, 2), (0, 2), (0, 3), (2, 1), (2, 0) \}, + \mod 4)$
$->$
$|Z_{2}^{+} \times Z_{4}^{+}| = 8$

so

• $1 ↔ (0, 0)$
• $3 ↔ (0, 1)$
• $5 ↔ (2, 3)$
• $7 ↔ (2, 2)$
• $9 ↔ (0, 2)$
• $11 ↔ (0, 3)$
• $13 ↔ (2, 1)$
• $15 ↔ (2, 0)$

Let's finally define the isomorphism

$(x \in Z_{\phi(16)}^{\ast} ↔ Z_{2}^{+} \times Z_{4}^{+}) + (y \in Z_{\phi(16)}^{\ast} ↔ Z_{2}^{+} \times Z_{4}^{+}) \mod 4 = (xy \in Z_{\phi(16)}^{\ast} ↔ Z_{2}^{+} \times Z_{4}^{+})$

Honestly I don't know if it is possible to derive a simpler/different isomorphism but my construction seems to work incredibly. Another interesting and strange behaviour of this kind of isomorphism is that it will hold for $Z_{\phi(15)}^{\ast}$ too, indeed this group is strangely correlated with $Z_{\phi(16)}^{\ast}$ since $\phi(15) = 2^{3}$ and $\phi(2^{4}) = 2^{3}$. Readapting the previous structures produces

$Z_{\phi(15)}^{\ast} = (\{ 1, 2, 4, 7, 8, 11, 13, 14 \}, \cdot \mod 15)$
$->$
$|Z_{\phi(16)}^{\ast}| = 8$

where $2$ produces a subgroup of order $4$ and $11$ is an independent generator. This means that we can start from them to build everything because as I already said, since $2$ produces $2$ generators for two other subgroups then $\{ 2 \cdot 11, 2^{2} \cdot 11, 2^{3} \cdot 11 \}$ will produce the remaining $3$ generators.

• $2 ↔ (0, 1)$
• $11 ↔ (2, 2)$
• $->$
• $2^2 \mod 15 = 4 ↔ (0, 1) + (0, 1) = (0, 2)$
• $2^3 \mod 15 = 8 ↔ (0, 2) + (0, 1) = (0, 3)$
• $2^4 \mod 15 = 1 ↔ (0, 3) + (0, 1) = (0, 0)$
• $and$
• $11^{2} \mod 15 = 1 ↔ (2, 2) + (2, 2) = (0, 0)$
• $then$
• $2 \cdot 11 \mod 15 = 7 ↔ (0, 1) + (2, 2) = (2, 3)$
• $2^{2} \cdot 11 \mod 15 = 14 ↔ (0, 2) + (2, 2) = (2, 0)$
• $2^{3} \cdot 11 \mod 15 = 13 ↔ (0, 3) + (2, 2) = (2, 1)$
• $also$
• $4^{2} \mod 15 = 1 ↔ (0, 2) + (0, 2) = (0, 0)$
• $7^{2} \mod 15 = 4 ↔ (2, 3) + (2, 3) = (0, 2)$
• $7^{3} \mod 15 = 13 ↔ (0, 2) + (2, 3) = (2, 1)$
• $7^{4} \mod 15 = 1 ↔ (2, 1) + (2, 3) = (0, 0)$
• $8^{2} \mod 15 = 4 ↔ (0, 3) + (0, 3) = (0, 2)$
• $8^{2} \mod 15 = 2 ↔ (0, 2) + (0, 3) = (0, 1)$
• $8^{2} \mod 15 = 1 ↔ (0, 3) + (0, 1) = (0, 0)$
• $9^{2} \mod 15 = 1 ↔ (0, 2) + (0, 2) = (0, 0)$
• $13^{2} \mod 15 = 4 ↔ (2, 1) + (2, 1) = (0, 2)$
• $13^{3} \mod 15 = 7 ↔ (0, 2) + (2, 1) = (2, 3)$
• $13^{4} \mod 15 = 1 ↔ (2, 3) + (2, 1) = (0, 0)$
• $14^{2} \mod 15 = 1 ↔ (2, 0) + (2, 0) = (0, 0)$

As you can see this construction is from a structural point of view the same we made for $Z_{\phi(16)}^{\ast}$, and we can define the isomorphism

$(x \in Z_{\phi(15)}^{\ast} ↔ Z_{2}^{+} \times Z_{4}^{+}) + (y \in Z_{\phi(15)}^{\ast} ↔ Z_{2}^{+} \times Z_{4}^{+}) \mod 4 = (xy \in Z_{\phi(15)}^{\ast} ↔ Z_{2}^{+} \times Z_{4}^{+})$

Generalizing we can observe that for $Z_{\phi(32)}^{\ast}$ etc. and groups having an equivalent $\phi$ it will be possible (always?) to construct similar isomorphisms since for $Z_{\phi(32)}^{\ast}$ etc., recalling [https://github.com/Z323323/Group-theory-elements?tab=readme-ov-file#theorem] we know we'll always have $2/3$ ($1$ will be useless) independent subgroups generators which we can use along with some subgroup generator of order $2^{k - 2}$ where $2^{k - 2} \equiv 1 \mod 2^{k}$ [ thus $\phi = 2^{k - 1}$ ] to generate the remaining $2^{k - 2}$ generators where $2^{k - 2} + 2^{k - 2} = 2^{k - 1} = \phi(n)$ and the related additive isomorphism.

To conclude, $Z_{\phi(15)}^{\ast} \cong Z_{4}^{+} \times Z_{2}^{+}$ means that $Z_{\phi(15)}^{\ast}$ is isomorphic to $Z_{4}^{+} \times Z_{2}^{+}$.

Let's now define the isomorphism $Z_{\phi(15)}^{\ast} \cong Z_{2}^{+} \times Z_{2}^{+} \times Z_{2}^{+}$. We transform the previous structure into

• $2 ↔ (\{ 0, 0 \}, \{ 0, 1 \})$
• $11 ↔ (\{ 1, 0 \}, \{ 1, 0 \})$
• $->$
• $2^2 \mod 15 = 4 ↔ (\{ 0, 0 \}, \{ 0, 1 \}) + (\{ 0, 0 \}, \{ 0, 1 \}) = (\{ 0, 0 \}, \{ 1, 0 \})$
• $2^3 \mod 15 = 8 ↔ (\{ 0, 0 \}, \{ 1, 0 \}) + (\{ 0, 0 \}, \{ 0, 1 \}) = (\{ 0, 0 \}, \{ 1, 1 \})$
• $2^4 \mod 15 = 1 ↔ (\{ 0, 0 \}, \{ 1, 1 \}) + (\{ 0, 0 \}, \{ 0, 1 \}) = (\{ 0, 0 \}, \{ 0, 0 \})$
• $and$
• $11^{2} \mod 15 = 1 ↔ (\{ 1, 0 \}, \{ 1, 0 \}) + (\{ 1, 0 \}, \{ 1, 0 \}) = (\{ 0, 0 \}, \{ 0, 0 \})$
• $then$
• $2 \cdot 11 \mod 15 = 7 ↔ (\{ 0, 0 \}, \{ 0, 1 \}) + (\{ 1, 0 \}, \{ 1, 0 \}) = (\{ 1, 0 \}, \{ 1, 1 \})$
• $2^{2} \cdot 11 \mod 15 = 14 ↔ (\{ 0, 0 \}, \{ 1, 0 \}) + (\{ 1, 0 \}, \{ 1, 0 \}) = (\{ 1, 0 \}, \{ 0, 0 \})$
• $2^{3} \cdot 11 \mod 15 = 13 ↔ (\{ 0, 0 \}, \{ 1, 1 \}) + (\{ 1, 0 \}, \{ 1, 0 \}) = (\{ 1, 0 \}, \{ 0, 1 \})$
• $also$
• $4^{2} \mod 15 = 1 ↔ (\{ 0, 0 \}, \{ 1, 0 \}) + (\{ 0, 0 \}, \{ 1, 0 \}) = (\{ 0, 0 \}, \{ 0, 0 \})$
• $7^{2} \mod 15 = 4 ↔ (\{ 1, 0 \}, \{ 1, 1 \}) + (\{ 1, 0 \}, \{ 1, 1 \}) = (\{ 0, 0 \}, \{ 1, 0 \})$
• $7^{3} \mod 15 = 13 ↔ (\{ 0, 0 \}, \{ 1, 0 \}) + (\{ 1, 0 \}, \{ 1, 1 \}) = (\{ 1, 0 \}, \{ 0, 1 \})$
• $7^{4} \mod 15 = 1 ↔ (\{ 1, 0 \}, \{ 0, 1 \}) + (\{ 1, 0 \}, \{ 1, 1 \}) = (\{ 0, 0 \}, \{ 0, 0 \})$
• $8^{2} \mod 15 = 4 ↔ (\{ 0, 0 \}, \{ 1, 1 \}) + (\{ 0, 0 \}, \{ 1, 1 \}) = (\{ 0, 0 \}, \{ 1, 0 \})$
• $8^{2} \mod 15 = 2 ↔ (\{ 0, 0 \}, \{ 1, 0 \}) + (\{ 0, 0 \}, \{ 1, 1 \}) = (\{ 0, 0 \}, \{ 0, 1 \})$
• $8^{2} \mod 15 = 1 ↔ (\{ 0, 0 \}, \{ 1, 1 \}) + (\{ 0, 0 \}, \{ 0, 1 \}) = (\{ 0, 0 \}, \{ 0, 0 \})$
• $9^{2} \mod 15 = 1 ↔ (\{ 0, 0 \}, \{ 1, 0 \}) + (\{ 0, 0 \}, \{ 1, 0 \}) = (\{ 0, 0 \}, \{ 0, 0 \})$
• $13^{2} \mod 15 = 4 ↔ (\{ 1, 0 \}, \{ 0, 1 \}) + (\{ 1, 0 \}, \{ 0, 1 \}) = (\{ 0, 0 \}, \{ 1, 0 \})$
• $13^{3} \mod 15 = 7 ↔ (\{ 0, 0 \}, \{ 1, 0 \}) + (\{ 1, 0 \}, \{ 0, 1 \}) = (\{ 1, 0 \}, \{ 1, 1 \})$
• $13^{4} \mod 15 = 1 ↔ (\{ 1, 0 \}, \{ 1, 1 \}) + (\{ 1, 0 \}, \{ 0, 1 \}) = (\{ 0, 0 \}, \{ 0, 0 \})$
• $14^{2} \mod 15 = 1 ↔ (\{ 1, 0 \}, \{ 0, 0 \}) + (\{ 1, 0 \}, \{ 0, 0 \}) = (\{ 0, 0 \}, \{ 0, 0 \})$