# Isomorphism of factors does not imply isomorphism of quotient groups

Let $$G$$ be a group and $$H, K$$ two isomorphic subgroups. We provide an example where the quotient groups $$G / H$$ and $$G / K$$ are not isomorphic.

Let $$G = \mathbb{Z}_4 \times \mathbb{Z}_2$$, with $$H = \langle (\overline{2}, \overline{0}) \rangle$$ and $$K = \langle (\overline{0}, \overline{1}) \rangle$$. We have $H \cong K \cong \mathbb{Z}_2.$ The left cosets of $$H$$ in $$G$$ are $G / H=\{(\overline{0}, \overline{0}) + H, (\overline{1}, \overline{0}) + H, (\overline{0}, \overline{1}) + H, (\overline{1}, \overline{1}) + H\},$ a group having $$4$$ elements and for all elements $$x \in G/H$$, one can verify that $$2x = H$$. Hence $$G / H \cong \mathbb{Z}_2 \times \mathbb{Z}_2$$. The left cosets of $$K$$ in $$G$$ are $G / K=\{(\overline{0}, \overline{0}) + K, (\overline{1}, \overline{0}) + K, (\overline{2}, \overline{0}) + K, (\overline{3}, \overline{0}) + K\},$ which is a cyclic group of order $$4$$ isomorphic to $$\mathbb{Z}_4$$. We finally get the desired conclusion $G / H \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \ncong \mathbb{Z}_4 \cong G / K.$

# A group that is not a semi-direct product

Given a group $$G$$ with identity element $$e$$, a subgroup $$H$$, and a normal subgroup $$N \trianglelefteq G$$; then we say that $$G$$ is the semi-direct product of $$N$$ and $$H$$ (written $$G=N \rtimes H$$) if $$G$$ is the product of subgroups, $$G = NH$$ where the subgroups have trivial intersection $$N \cap H= \{e\}$$.

Semi-direct products of groups provide examples of non abelian groups. For example the dihedral group $$D_{2n}$$ with $$2n$$ elements is isomorphic to a semidirect product of the cyclic groups $$\mathbb Z_n$$ and $$\mathbb Z_2$$. $$D_{2n}$$ is the group of isometries preserving a regular polygon $$X$$ with $$n$$ edges.

Let’see that the converse is not true and present a group that is not a semi-direct product.

### The Hamilton’s quaternions group is not a semi-direct product

The Hamilton’s quaternions group $$\mathbb H_8$$ is the group consisting of the symbols $$\pm 1, \pm i, \pm j, \pm k$$ where$-1 = i^2 =j^2 = k^2 \text{ and } ij = k = -ji,jk = i = -kj, ki = j = -ik.$ One can prove that $$\mathbb H_8$$ endowed with the product operation above is indeed a group having $$8$$ elements where $$1$$ is the identity element.

$$\mathbb H_8$$ is not abelian as $$ij = k \neq -k = ji$$.

Let’s prove that $$\mathbb H_8$$ is not the semi-direct product of two subgroups. If that was the case, there would exist a normal subgroup $$N$$ and a subgroup $$H$$ such that $$G=N \rtimes H$$.

• If $$\vert N \vert = 4$$ then $$H = \{1,h\}$$ where $$h$$ is an element of order $$2$$ in $$\mathbb H_8$$. Therefore $$h=-1$$ which is the only element of order $$2$$. But $$-1 \in N$$ as $$-1$$ is the square of all elements in $$\mathbb H_8 \setminus \{\pm 1\}$$. We get the contradiction $$N \cap H \neq \{1\}$$.
• If $$\vert N \vert = 2$$ then $$\vert H \vert = 4$$ and $$H$$ is also normal in $$G$$. Noting $$N=\{1,n\}$$ we have for $$h \in H$$ $$h^{-1}nh=n$$ and therefore $$nh=hn$$. This proves that the product $$G=NH$$ is direct. Also $$N$$ is abelian as a cyclic group of order $$2$$. $$H$$ is also cyclic as all groups of order $$p^2$$ with $$p$$ prime are abelian. Finally $$G$$ would be abelian, again a contradiction.

We can conclude that $$G$$ is not a semi-direct product.

# A normal subgroup that is not a characteristic

Let’s $$G$$ be a group. A characteristic subgroup is a subgroup $$H \subseteq G$$ that is mapped to itself by every automorphism of $$G$$.

An inner automorphism is an automorphism $$\varphi \in \mathrm{Aut}(G)$$ defined by a formula $$\varphi : x \mapsto a^{-1}xa$$ where $$a$$ is an element of $$G$$. An automorphism of a group which is not inner is called an outer automorphism. And a subgroup $$H \subseteq G$$ that is mapped to itself by every inner automorphism of $$G$$ is called a normal subgroup.

Obviously a characteristic subgroup is a normal subgroup. The converse is not true as we’ll see below.

### Example of a direct product

Let $$K$$ be a nontrivial group. Then consider the group $$G = K \times K$$. The subgroups $$K_1=\{e\} \times K$$ and $$K_2=K \times \{e\}$$ are both normal in $$G$$ as for $$(e, k) \in K_1$$ and $$(a,b) \in G$$ we have
$(a,b)^{-1} (e,x) (a,b) = (a^{-1},b^{-1}) (e,x) (a,b) = (e,b^{-1}xb) \in K_1$ and $$b^{-1}K_1 b = K_1$$. Similar relations hold for $$K_2$$. As $$K$$ is supposed to be nontrivial, we have $$K_1 \neq K_2$$.

The exchange automorphism $$\psi : (x,y) \mapsto (y,x)$$ exchanges the subgroup $$K_1$$ and $$K_2$$. Thus, neither $$K_1$$ nor $$K_2$$ is invariant under all the automorphisms, so neither is characteristic. Therefore, $$K_1$$ and $$K_2$$ are both normal subgroups of $$G$$ that are not characteristic.

When $$K = \mathbb Z_2$$ is the cyclic group of order two, $$G = \mathbb Z_2 \times \mathbb Z_2$$ is the Klein four-group. In particular, this gives a counterexample where the ambient group is an abelian group.

### Example on the additive group $$\mathbb Q$$

Consider the additive group $$(\mathbb Q,+)$$ of rational numbers. The map $$\varphi : x \mapsto x/2$$ is an automorphism. As $$(\mathbb Q,+)$$ is abelian, all subgroups are normal. However, the subgroup $$\mathbb Z$$ is not sent into itself by $$\varphi$$ as $$\varphi(1) = 1/ 2 \notin \mathbb Z$$. Hence $$\mathbb Z$$ is not a characteristic subgroup.

# Is the quotient group of a finite group always isomorphic to a subgroup?

Given a normal subgroup $$H$$ of a finite group $$G$$, is $$G/H$$ always isomorphic to a subgroup $$K \le G$$?

### The case of an abelian group

According to the fundamental theorem of finite abelian groups, every finite abelian group $$G$$ can be expressed as the direct sum of cyclic subgroups of prime-power order: $G \simeq \bigoplus_{i=1}^u \mathbb{Z}_{p_i^{\alpha_i}}$ where $$p_1, \dots , p_u$$ are primes and $$\alpha_1, \dots , \alpha_u$$ non zero integers.

If $$H \le G$$ we have $H \simeq \bigoplus_{i=1}^u \mathbb{Z}_{p_i^{\beta_i}}$ with $$0 \le \beta_1 \le \alpha_1, \dots, 0 \le \beta_u \le \alpha_u$$. Then $G/H \simeq \bigoplus_{i=1}^u \mathbb{Z}_{p_i^{\alpha_i-\beta_i}}$ which is a subgroup of $$G$$.

If $$G$$ is not abelian, then $$G/H$$ might not be isomorphic to a subgroup of $$G$$. Continue reading Is the quotient group of a finite group always isomorphic to a subgroup?

# A simple group whose order is not a prime

Consider a finite group $$G$$ whose order (number of elements) is a prime number. It is well known that $$G$$ is cyclic and simple. Which means that $$G$$ has no non trivial normal subgroup.

Is the converse true, i.e. are the cyclic groups with prime orders the only simple groups? The answer is negative. We prove here that for $$n \ge 5$$ the alternating group $$A_n$$ is simple. In particular $$A_5$$ whose order is equal to $$60$$ is simple. Continue reading A simple group whose order is not a prime

# An infinite group whose proper subgroups are all finite

We study some properties of the Prüfer $$p$$-group $$\mathbb{Z}_{p^\infty}$$ for a prime number $$p$$. The Prüfer $$p$$-group may be identified with the subgroup of the circle group, consisting of all $$p^n$$-th roots of unity as $$n$$ ranges over all non-negative integers:
$\mathbb{Z}_{p^\infty}=\bigcup_{k=0}^\infty \mathbb{Z}_{p^k} \text{ where } \mathbb{Z}_{p^k}= \{e^{\frac{2 i \pi m}{p^k}} \ | \ 0 \le m \le p^k-1\}$

### $$\mathbb{Z}_{p^\infty}$$ is a group

First, let’s notice that for $$0 \le m \le n$$ integers we have $$\mathbb{Z}_{p^m} \subseteq \mathbb{Z}_{p^n}$$ as $$p^m | p^n$$. Also for $$m \ge 0$$ $$\mathbb{Z}_{p^m}$$ is a subgroup of the circle group. We also notice that all elements of $$\mathbb{Z}_{p^\infty}$$ have finite orders which are powers of $$p$$. Continue reading An infinite group whose proper subgroups are all finite

# The set of all commutators in a group need not be a subgroup

I here provide a simple example of a group whose set of commutators is not a subgroup. The counterexample is due to P.J. Cassidy (1979).

### Description of the group $$G$$

Let $$k[x,y]$$ denote the ring of all polynomials in two variables over a field $$k$$, and let $$k[x]$$ and $$k[y]$$ denote the subrings of all polynomials in $$x$$ and in $$y$$ respectively. $$G$$ is the set of all upper unitriangular matrices of the form
$A=\left(\begin{array}{ccc} 1 & f(x) & h(x,y) \\ 0 & 1 & g(y) \\ 0 & 0 & 1 \end{array}\right)$ where $$f(x) \in k[x]$$, $$g(y) \in k[y]$$, and $$h(x,y) \in k[x,y]$$. The matrix $$A$$ will also be denoted $$(f,g,h)$$.
Let’s verify that $$G$$ is a group. The products of two elements $$(f,g,h)$$ and $$(f^\prime,g^\prime,h^\prime)$$ is
$\left(\begin{array}{ccc} 1 & f(x) & h(x,y) \\ 0 & 1 & g(y) \\ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{ccc} 1 & f^\prime(x) & h^\prime(x,y) \\ 0 & 1 & g^\prime(y) \\ 0 & 0 & 1 \end{array}\right)$
$=\left(\begin{array}{ccc} 1 & f(x)+f^\prime(x) & h(x,y)+h^\prime(x,y)+f(x)g^\prime(y) \\ 0 & 1 & g(y)+g^\prime(y) \\ 0 & 0 & 1 \end{array}\right)$ which is an element of $$G$$. We also have:
$\left(\begin{array}{ccc} 1 & f(x) & h(x,y) \\ 0 & 1 & g(y) \\ 0 & 0 & 1 \end{array}\right)^{-1} = \left(\begin{array}{ccc} 1 & -f(x) & f(x)g(y) – h(x,y) \\ 0 & 1 & -g(y) \\ 0 & 0 & 1 \end{array}\right)$ proving that the inverse of an element of $$G$$ is also an element of $$G$$. Continue reading The set of all commutators in a group need not be a subgroup

# Two subgroups whose product is not a subgroup

In this article, we consider a group $$G$$ and two subgroups $$H$$ and $$K$$. Let $$HK=\{hk \text{ | } h \in H, k \in K\}$$.

$$HK$$ is a subgroup of $$G$$ if and only if $$HK=KH$$. For the proof we first notice that if $$HK$$ is a subgroup of $$G$$ then it’s closed under inverses so $$HK = (HK)^{-1} = K^{-1}H^{-1} = KH$$. Conversely if $$HK = KH$$ then take $$hk$$, $$h^\prime k^\prime \in HK$$. Then $$(hk)(h^\prime k^\prime)^{-1} = hk(k^\prime)^{-1}(h^\prime)^{-1}$$. Since $$HK = KH$$ we can rewrite $$k(k^\prime)^{-1}(h^\prime)^{-1}$$ as $$h^{\prime \prime}k^{\prime \prime}$$ for some new $$h^{\prime \prime} \in H$$, $$k^{\prime \prime} \in K$$. So $$(hk)(h^\prime k^\prime)^{-1}=hh^{\prime \prime}k^{\prime \prime}$$ which is in $$HK$$. This verifies that $$HK$$ is a subgroup. Continue reading Two subgroups whose product is not a subgroup

# A finitely generated soluble group isomorphic to a proper quotient group

Let $$\mathbb{Q}_2$$ be the ring of rational numbers of the form $$m2^n$$ with $$m, n \in \mathbb{Z}$$ and $$N = U(3, \mathbb{Q}_2)$$ the group of unitriangular matrices of dimension $$3$$ over $$\mathbb{Q}_2$$. Let $$t$$ be the diagonal matrix with diagonal entries: $$1, 2, 1$$ and put $$H = \langle t, N \rangle$$. We will prove that $$H$$ is finitely generated and that one of its quotient group $$G$$ is isomorphic to a proper quotient group of $$G$$. Continue reading A finitely generated soluble group isomorphic to a proper quotient group

# Converse of Lagrange’s theorem does not hold

Lagrange’s theorem, states that for any finite group $$G$$, the order (number of elements) of every subgroup $$H$$ of $$G$$ divides the order of $$G$$ (denoted by $$\vert G \vert$$).

Lagrange’s theorem raises the converse question as to whether every divisor $$d$$ of the order of a group is the order of some subgroup. According to Cauchy’s theorem this is true when $$d$$ is a prime.

However, this does not hold in general: given a finite group $$G$$ and a divisor $$d$$ of $$\vert G \vert$$, there does not necessarily exist a subgroup of $$G$$ with order $$d$$. The alternating group $$G = A_4$$, which has $$12$$ elements has no subgroup of order $$6$$. We prove it below. Continue reading Converse of Lagrange’s theorem does not hold