A nonabelian $$p$$-group

Consider a prime number $$p$$ and a finite p-group $$G$$, i.e. a group of order $$p^n$$ with $$n \ge 1$$.

If $$n=1$$ the group $$G$$ is cyclic hence abelian.

For $$n=2$$, $$G$$ is also abelian. This is a consequence of the fact that the center $$Z(G)$$ of a $$p$$-group is non-trivial. Indeed if $$\vert Z(G) \vert =p^2$$ then $$G=Z(G)$$ is abelian. We can’t have $$\vert Z(G) \vert =p$$. If that would be the case, the order of $$H=G / Z(G)$$ would be equal to $$p$$ and $$H$$ would be cyclic, generated by an element $$h$$. For any two elements $$g_1,g_2 \in G$$, we would be able to write $$g_1=h^{n_1} z_1$$ and $$g_2=h^{n_1} z_1$$ with $$z_1,z_2 \in Z(G)$$. Hence $g_1 g_2 = h^{n_1} z_1 h^{n_2} z_2=h^{n_1 + n_2} z_1 z_2= h^{n_2} z_2 h^{n_1} z_1=g_2 g_1,$ proving that $$g_1,g_2$$ commutes in contradiction with $$\vert Z(G) \vert < \vert G \vert$$. However, all $$p$$-groups are not abelian. For example the unitriangular matrix group $U(3,\mathbb Z_p) = \left\{ \begin{pmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1\end{pmatrix} \ | \ a,b ,c \in \mathbb Z_p \right\}$ is a $$p$$-group of order $$p^3$$. Its center $$Z(U(3,\mathbb Z_p))$$ is $Z(U(3,\mathbb Z_p)) = \left\{ \begin{pmatrix} 1 & 0 & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix} \ | \ b \in \mathbb Z_p \right\},$ which is of order $$p$$. Therefore $$U(3,\mathbb Z_p)$$ is not abelian.

A group isomorphic to its automorphism group

We consider a group $$G$$ and we look at its automorphism group $$\text{Aut}(G)$$. Can $$G$$ be isomorphic to
$$\text{Aut}(G)$$?
The answer is positive and we’ll prove that it is the case for the symmetric group $$S_3$$.

Consider the morphism $\begin{array}{l|rcl} \Phi : & S_3 & \longrightarrow & \text{Aut}(S_3) \\ & a & \longmapsto & \varphi_a \end{array}$
where $$\varphi_a$$ is the inner automorphism $$\varphi_a : x \mapsto a^{-1}xa$$. It is easy to verify that $$\Phi$$ is indeed a group morphism. The kernel of $$\Phi$$ is the center of $$S_3$$ which is having the identity for only element. Hence $$\Phi$$ is one-to-one and $$S_3 \simeq \Phi(S_3)$$. Therefore it is sufficient to prove that $$\Phi$$ is onto. As $$|S_3|=6$$, we’ll be finished if we prove that $$|\text{Aut}(S_3)|=6$$.

Generally, for $$G_1,G_2$$ groups and $$f : G_1 \to G_2$$ a one-to-one group morphism, the image of an element $$x$$ of order $$k$$ is an element $$f(x)$$ having the same order $$k$$. So for $$\varphi \in \text{Aut}(S_3)$$ the image of a transposition is a transposition. As the transpositions $$\{(1 \ 2), (1 \ 3), (2 \ 3)\}$$ generate $$(S_3)$$, $$\varphi$$ is completely defined by $$\{\varphi((1 \ 2)), \varphi((1 \ 3)), \varphi((2 \ 3))\}$$. We have 3 choices to define the image of $$(1 \ 2)$$ under $$\varphi$$ and then 2 choices for the image of $$(1 \ 3)$$ under $$\varphi$$. The image of $$(2 \ 3)$$ under $$\varphi$$ is the remaining transposition.

Finally, we have proven that $$|\text{Aut}(S_3)|=6$$ as desired and $$S_3 \simeq \text{Aut}(S_3)$$.

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

Generating the symmetric group with a transposition and a maximal length cycle

Can the symmetric group $$\mathcal{S}_n$$ be generated by any transposition and any $$n$$-cycle for $$n \ge 2$$ integer? is the question we deal with.

We first recall some terminology:

Symmetric group
The symmetric group $$\mathcal{S}_n$$ on a finite set of $$n$$ symbols is the group whose elements are all the permutations of the $$n$$ symbols. We’ll denote by $$\{1,\dots,n\}$$ those $$n$$ symbols.
Cycle
A cycle of length $$k$$ (with $$k \ge 2$$) is a cyclic permutation $$\sigma$$ for which there exists an element $$i \in \{1,\dots,n\}$$ such that $$i, \sigma(i), \sigma^2(i), \dots, \sigma^k(i)=i$$ are the only elements moved by $$\sigma$$. We’ll denote the cycle $$\sigma$$ by $$(s_0 \ s_1 \dots \ s_{k-1})$$ where $$s_0=i, s_1=\sigma(i),\dots,s_{k-1}=\sigma^{k-1}(i)$$.
Transposition
A transposition is a cycle of length $$2$$. We denote below the transposition of elements $$a \neq b$$ by $$(a \ b)$$ or $$\tau_{a,b}$$.

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

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