Prove that a group a non-abelian of order 6 is isomorphic to $S_3$. Prove that every abelian group of order 6 is isomorphic to $Z/{6Z}$. Here are some hints: start by showing that every group $G$ of order 6 must have an element $x$ of order 2 and an element $y$ of order 3. This in fact follows from some general theorems but I want you to argue directly using only what we covered in class. (A typical problem here is why can’t all the elements different from 1 have order 3. If this is the case, show that there are two cyclic groups $K_1,K_2$ of $G$ of order 3 such that $K_1 \cap K_2 = \left\{1\right\}$. Calculate $|K_1K_2|$.) Having shown that, if $G$ is abelian show it implies the existence of an element of order 6. In the non-abelian case show that we must have $xyx^{-1} = y^2$ and that every element in $G$ is of the form $x^ay^b$, $a = 0, 1, b = 0, 1, 2$. Show that the map $x\to (1 2)$, $y\to (1 2 3)$ extends to an isomorphism.

Hi. I am trying to prove the hint. But I cannot conclude that the group has an element of order 2 and one of order 3. I have the following:

My Solution. Suppose $\forall g\in G,\ g\neq e,\ |g|=3$. Let $g,h\in G,\ g\neq h$. Then $\langle g\rangle\cap \langle h\rangle=\left\{e\right\}$. Indeed, $\langle g\rangle=\left\{e,g,g^2\right\} and \langle h\rangle=\left\{e,h,h^2\right\}$. If $g=h^2\Rightarrow gh=h^3=e\Rightarrow h=g^{-1}\wedge h^2=g^{-2}\Rightarrow \langle h\rangle=\left\{e,g^{-1},g^{-2}\right\}=\left\{e,g,g^2\right\}$.

In general, if $G=\left\{e,g_1,g_2,g_3,g_4,g_5\right\}$ then $\langle g_i\rangle=\langle g_1\rangle,\ i=2,3,4,5$.

Now, $G=\bigcup_{i=1}^{5} \langle g_i\rangle=\left\{e,g_1^2,g_1^2\right\}$ a contradiccion with $|G|=6$.

Now, $|\langle g\rangle \langle h\rangle|=9$ a contradiction with $|G|=6.$

Therefore, exists $g\in G,\ g\neq e$ such that $|g|\in \left\{2,6\right\}$.

Now, can’t all the elements different from 1 have order 2. Suppose that for all $g\in G,g\neq e,\ |g|=2 \Rightarrow G$ abelian $\Rightarrow S=\left\{e,g,h,gh\right\}$ subgroup of $G$ but $|S|\not\mid |G|$ a contradiction.

Therefore, exists $g\in G,\ g\neq e,\ |g|\in\left\{3,6\right\}$.

Why exists $x,y\in G$ such that $|x|=2, |y|=3$?

Actualization 1. I ahve proves this exists $x,y\in G$ such that $|x|\in \left\{2,6\right\}$ and $|y|\in\left\{3,6\right\}$.

If $|x|=6$ then $|x^3|=2$ and $|x^2|=3$. Therefore $x^3, x^2$ are elements in $G$ of order 2 and 3 respectively.

If $|x|=2$ then x is the element of order 2. If $|y|=6$ similary. If $|y|=3$ then $x,y$ are elements of order 2 and 3.

Now, if $G$ abelian $|(xy)|=6$ then $G\simeq Z_6$. If $G$ no abelian. How proves that $xyx^{-1}=y^2$?

Actualization 2. Let $G$ non aebelian. $ [G:\langle y\rangle ]=2$ then $\langle y\rangle$ normal in $G$ then $x\langle y\rangle=\langle y\rangle$. therefore $xyx^{-1}\in \left\{e,y,y^2\right\}$ then $xyx^{-1}=y^2$ (other cases are contradiction)


2 Answers 2


For showing that there are elements of order 2 and 3, let's first look into the possibility of orders of nonidentity elements which are 2,3,6.

If you have an element of order 6, you are done (why?).

Otherwise suppose all the elements($\neq 1$) are of order 2 then $G$ must be abelian. In that case $G$ is a $\mathbb Z/2$ vector space of finite dimension hence of cardinality $2^k$ for some k>0, contradiction.

So, if your group has an order 2 element it must have one order 3 element.

Now suppose it has all the nonidentity elements of order 3 Then write elements, $\{1, a, a^2\}$ next choose some element $b$ outside this collection argue that $\{b, b^2\}$ is disjoint from the above collection. Now there will still be another nonidentity element let's say $c$ in G. Show that, $c^2 \notin \{1, a, a^2, b, b^2,c\}$. Hence a contradiction.

  • $\begingroup$ But how could you prove that you could have an element of order 2 and 3 without distinguishing the cases (G has an element of order 6, G does not have an element of order 6, G is abelian, G is not abelian, etc., etc.) proves it directly. $\endgroup$
    – eraldcoil
    May 2, 2020 at 16:56
  • $\begingroup$ This is proving it directly, isn't it? $\endgroup$ May 2, 2020 at 18:35

You can use sylow theorem. According to $6 = 2*3$ and that every prime order group is cyclical group, we can easily prove the question in your title.

  • 1
    $\begingroup$ If Cauchy can't be brought in, I guess Sylow is even less so. $\endgroup$
    – user750041
    May 2, 2020 at 19:12

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