Prove that every non-abelian group of order $6$ has a non-normal subgroup of order $2$. Use this to classify groups of order $6$.

I proved that every non-abelian group of order 6 has a nonnormal subgroup of order 2, but then how can I use this to classify groups of order 6?

  • $\begingroup$ can we assume you know group actions??? $\endgroup$ – user87543 Sep 17 '13 at 5:34
  • $\begingroup$ yes~ @PraphullaKoushik I just learned it.. $\endgroup$ – Tumbleweed Sep 17 '13 at 5:40
  • $\begingroup$ then, that would be easy :) $\endgroup$ – user87543 Sep 17 '13 at 5:44

Assuming $H$ is a non normal subgroup of order $2$.

Consider Action of $G$ on set of left cosets of $H$ by left multiplication.

let $\{g_iH :1\leq i\leq 3\}$ be cosets of $H$ in $G$.

(please convince yourself that there will be three distinct cosets)

we now consider the action $\eta : G\times\{g_iH :1\leq i\leq 3\} \rightarrow \{g_iH :1\leq i\leq 3\}$ by left multiplication.

i.e., take an element $g\in G$ and consider $g.g_iH$ as there are only three distinct sets we get $g.g_iH = g_jH$ for some $j\in \{1,2,3\}$

In this manner, the elements of $g\in G$ takes cosets $g_iH$ (represented by $i$) to cosets $g_jH$ (represented by $j$)

i.e., we have map $\eta :\{1,2,3\} \rightarrow \{1,2,3\}$ which can be seen as $\eta : G\rightarrow S_3$

we know that $Ker(\eta)$ is normal in $G$ which is contained in $H$.

As $H$ is not normal in $G$ we end up with the case that $Ker(\eta)=(1)$ i.e., $\eta$ is injective.

i.e., we have $G$ as a subgroup(isomorphic copy) of $S_3$. But, $|G|=|S_3|=6$. Thus, $G\cong S_3$.

So, for any non abelian group $G$ of order $6$ we have $G\cong S_3$.

For an abelian group of order $6$ we have already know that $G$ is cyclic and $G\cong \mathbb{Z}_6$.

So, only non isomorphic groups of order $6$ are $S_3,\mathbb{Z}_6$.

  • $\begingroup$ Thank you Prapulla. It is not obvious to me that $\ker(\eta)$ is normal in $G$ which is contained in $H$, and $\ker(\eta) = (1)$. Could you explain...? $\endgroup$ – Tumbleweed Sep 17 '13 at 5:57
  • $\begingroup$ okay, let us first check why $ker(\eta)$ is normal... where does $gkg^{-1}$ go for $k\in Ker(\eta)$ and any $g\in G$??? i mean what is $\eta(gkg^{-1})$... ??? USE that $\eta$ is a "homomorphism".... $\endgroup$ – user87543 Sep 17 '13 at 6:00
  • $\begingroup$ Hmm, interesting, thanks for your question: $$\eta(gkg^{-1}) = \eta(g) \eta(k) \eta(g^{-1}) = \eta(g)\cdot id \cdot \eta(g^{-1}) = \eta(g)\eta(g^{-1}) = \eta(gg^{-1}) = \eta(id) = id.$$ $\endgroup$ – Tumbleweed Sep 17 '13 at 6:04
  • $\begingroup$ SO, you got that $ker(\eta)\unlhd G$????? $\endgroup$ – user87543 Sep 17 '13 at 6:05
  • $\begingroup$ Yes~~ Praphulla $\endgroup$ – Tumbleweed Sep 17 '13 at 6:09

Hint: You know that $G$ has an element of order $3$, this obviously generates a normal subgroup $H$ (why? think index). But, if it then had another subgroup $K$ of order $2$ which is also normal, then $G=HK$, $H\cap K=\{1\}$, and $H,K\unlhd G$. Why is that bad?

For the second part, let $G$ act on something that gives you an injective homomorphism $G\to S_3$. What should it be?

Just for fun, there is a much simpler way to do this. Note that if $H,K$ are above, then $HK=G$, $H\cap K=\{1\}$, and $H\unlhd G$. Thus, $G\cong H\rtimes_\varphi K$. But, note that $\text{Aut}(H)$ has order $2$, and so there are at most two distinct groups of order $6$. Since $\mathbb{Z}/6\mathbb{Z}$ and $S_3$ are distinct, they must constitute all of them. This acdtually shows you how to classify groups of order $2p$, $p$ an odd prime.


Let $G$ be a non-abelian group of order six. Then $G$ has exactly $3$ $2$-Sylow subgroups, which map to each other under conjugation. Denoting these Sylow subgroups $P_1,P_2,P_3$, we obtain a map (the permutation representation of the group action) $$\varphi: G \to \mathrm{Sym}(P_1,P_2,P_3)\cong S_3,$$ in which $\varphi(g)$ denotes the permutation of $P_1,P_2,P_3$ attained by conjugating each (respectively) by $g$. This map is injective, hence an isomorphism by orders. That is, $G \cong S_3$.

For the abelian case, it suffices to prove that such a group $G$ has an element of order $6$. To do this, it suffices to find elements of orders $2$ and $3$, which exist by a theorem of Cauchy (or later, by the Sylow theorems).

  • $\begingroup$ Can you explain why the map $\varphi (g)$ is injective? I know it cannot be trivial since all of the 3 2-Sylow subgroups are conjugate, but why can't it have a Kernel of size 2 or 3? $\endgroup$ – Dean Gurvitz Sep 15 '18 at 16:25

The following avoids Sylow theorems.

Let $G$ be a group of order $6$. Then $G$ has an element $a$ of order $3$ and an element $b$ of order $2$. Then $G=\langle a,b\rangle$ as the order of $\langle a,b\rangle$ is a multiple of both $3$ and $2$.

If $G$ is abelian, we conclude that $G\cong \mathbb Z/3\mathbb Z\oplus\mathbb Z/2\mathbb Z\cong\mathbb Z/6\mathbb Z$.

If $G$ is nonabelian. Let $A=\{b,aba^{-1},a^2ba^{-2}\}$. If two of the listed elements coincide (i.e. $|A|<3$) then $a$ normalizes $\langle b\rangle$ and since trivially $b$ normalizes $\langle b\rangle$, we find $\langle b\rangle\lhd G$. By what you have shown, we can pick $b$ such that this is not the case, i.e. such that $|A|=3$. Now we have one element of order $1$ in $G$, at least two elements of order $3$ and at least the three lements of $A$ of order $2$. We conclude that $A$ is precisely the set of elements of order $2$. Especially, these are conjugates and hence none of the subgroups of order two is normal. The group $G$ acts on $A$ by conjugation, which gives us a homomorphism $\phi\colon G\to S_3$. The kernel of $\phi$ is normal in $G$, hence cannot be of order $2$. It also does not contain $a$, hence it is trivial, i.e. $\phi$ is anisomorphism.


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