# $A_4$ has no subgroup of order $6$?

Can a kind algebraist offer an improvement to this sketch of a proof?

Show that $A_4$ has no subgroup of order 6.

Note, $|A_4|= 4!/2 =12$.
Suppose $A_4>H, |H|=6$.
Then $|A_4/H| = [A_4:H]=2$.
So $H \vartriangleleft A_4$ so consider the homomorphism
$\pi : A_4 \rightarrow A_4/H$
let $x \in A_4$ with $|x|=3$ (i.e. in a 3-cycle)
then 3 divides $|\pi(x)|$
so as $|A_4/H|=2$ we have $|\pi(x)|$ divides 2
so $\pi(x) = e_H$ so $x \in H$
so $H$ contains all 3-cycles
but $A_4$ has $8$ $3$-cycles
$8>6$, $A_4$ has no subgroup of order 6.

• I don't understand why this post is so heavily downvoted. The OP obviously tried. His proof may not be perfect, but if it were what would be the point of asking the question? Nov 27, 2013 at 2:16
• There's a typo: Where you wrote that 3 divides $|\pi(x)|$, you presumably meant that $|\pi(x)|$ divides 3. Apart from that, your proof looks OK. Dec 15, 2016 at 5:05
• I've also downvoted this question. The OP tried nothing: just copied a proof from somewhere and offered no clue on which part of this they are stuck. Dec 15, 2016 at 7:51

Consider the group $A_4/H$. Let $x$ be a $3$-cycle, not in $H$, and consider the cosets $H$, $xH$, and $x^2H$ in $A_4/H$. Since this is a group of order $2$, two of the cosets must be equal. But $H$ and $xH$ are distinct, so $x^2H$ must be equal to one of them.

If $H=x^2H$, then $x^2=x^{-1}\in H$, so $x\in H$, contradiction. If $xH=x^2H$, then $x\in H$, same problem. So $H$ doesn't exist.

• This is an old post so sorry for making a comment now, but how can you assure that there exists a $3-$cycle not contained in $H$? Sep 23, 2014 at 4:44
• @user156441 It's a well known fact that for any $n\geq 3$, $A_n$ is generated by $3$-cycles (here is a proof). So if $H$ contained all $3$-cycles, it would be $A_4$. Sep 23, 2014 at 5:09
• I'll take a look at the proof, nice answer by the way, thanks! Sep 23, 2014 at 5:10
• an other way : $\mathcal A_4$ has 8 as number of it's 3-cycles. Dec 1, 2014 at 1:37
• @E_K Right, $x^3=1$ as $x$ is a 3-cycle, so $x^2=x^{-1}$. If $H=x^2H$, then $x^2\in x^2H=H$, so $x^2=x^{-1}\in H$, and thus $x\in H$ since $H$ is closed under inverses as a subgroup. If $xH=x^2H$, then $H=x^{-1}xH=x^{-1}x^2H=xH$, so $x\in xH=H$. Sep 24, 2017 at 21:47

By looking at the possible cycle types, we see that $A_4$ consists of the identity element (order $1$), $3$ double transpositions (order $2$) and $8$ $3$-cycles (order $3$).

Assume that $A_4$ has a subgroup $H$ of order $6$. Since $A_4$ does not contain elements of order $6$, $H$ cannot be cyclic. Therefore $H \cong S_3$, implying that $H$ contains $3$ elements of order $2$. So $H$ contains the identity element and the $3$ double transpositions. Since those $4$ elements form a subgroup of $A_4$, $H$ contains a subgroup of order $4$. Contradiction.

I know this post is old, but there's another elegant way to prove this - a subgroup of order 6 has index 2. We prove the following statement: Any subgroup of index 2 of a finite group must contain all elements of odd order.

Let $G$ be finite and $H\subseteq G$ a subgroup of index 2. Any subgroup of index 2 is normal, so $G/H$ is a group and we write $\bar g:=gH$. Let $g$ be an element of odd order. Now we have $$g^{\operatorname{ord}g}=e\ \Rightarrow\ \bar g ^{\operatorname{ord}g}=\bar e\ \Rightarrow\ \operatorname{ord}\bar g\mid\operatorname{ord}g.$$ On the other hand, by Lagrange's theorem, we know that $\operatorname{ord}G/H =2$ so $$\bar g^2=\bar e\ \Rightarrow\ \operatorname{ord}\bar g\mid2.$$ Since $\operatorname{ord}g$ is odd, it follows that $$\operatorname{ord}\bar g=1\ \Rightarrow\ \bar g=\bar e=e_{G/H}=H\ \Rightarrow\ g\in H.$$

Now since $A_4$ contains 9 elements of odd order, a subgroup of index 2 would, by the above statement have at least 9 elements, but by Lagrange's theorem has exactly 6 elements, which is a contradiction.

• Excellent argument! Aug 16, 2017 at 17:27
• @Sora: From your profile I found that you have "an Introduction to LaTeX for mathematicians". I am thinking to learn LaTeX by my-self. Would you like to share it with me? Nov 27, 2017 at 4:16
• @Bumblebee I used to teach LaTeX at university - the lecture slides that I have are in German so I guess they won't be particularly useful... :/ but there are tons of LaTeX tutorials on the internet! Nov 27, 2017 at 4:18
• Tank you for your prompt reply. Yes, tutorials I tried were not suitable for a beginner like me. Any way thank you. Nov 27, 2017 at 4:21

I'm not sure if this is a combination of what people did above or not, but here's an approach that should work.

For the purpose of contradiction, assume $H\subset A_4$ is a subgroup of $A_4$ of order 6. Then, for any $a\notin H$ by properties of left cosets, then $aH\cap H=\emptyset$. Again, by properties of cosets, since $|aH|=|eH|=|H|$ (all cosets have the same number of elements), this implies that $|aH|=6$. Then, as cosets form a partition of the group $A_4$, and $|A_4|=12$, then $$A_4=H\cup aH$$ Now suppose that $a$ is a 3-cycle in $A_4$, then either $a^2\in H$ or $a^2\in aH$. If $a^2\in H$, then this implies that $a^4=a^2\cdot a^2\in H$ but, since the order of $a$ is 3 (it is a 3-cycle), then $a^4=a$ and since $a\notin H$ this is a contradiction. Similarly, if $a^2\in aH$ by properties of cosets this implies that $(a^2)\cdot a^{-1}\in H$ which implies $a\in H$ and again this is a contradiction.

As such, H cannot be a subgroup of $A_4$ of order 6.

I am a novice, just start to preview the content in the college. The following is my solution. If I make any mistakes, please point it out.

First, as a group of order 6 is either $$C_6$$ or $$D_6$$, we can merely discuss these two cases. Moreover, an even permutation of 4 elements can only be of the form (abc)(d) or (ab)(cd)

Case1: $$C_6$$. This is, of course, impossible as the maximum order is 4.

Case2: $$D_6$$. It is of the form {$$e,r,r^2,s,sr,sr^2$$}. r must of the form (abc)(d), and w.l.o.g., we can assume s of the form (ad)(bc).

Then it us easy to calculate sr=(acd), $$sr^2$$=(abd). However, in this case, (abd)^2=(adb), which is a new element outside of the six. So contradiction.

Based on reflections, $A_4$ is isometric to the rotation group of the tetrahedron. The tetrahedron has 4 vertices, so 4 subgroups of order 3. There are also 3 pairs of nonadjacent edges. So 3 subgroups of order 2. This exhausts all 12 elements of the group.

• "isometric"?!?!
– Pedro
Oct 15, 2014 at 21:04
• I don't see how this answers the question. Jun 21, 2015 at 18:15

Here's another way to look at this. It is well known that A4 has a normal subgroup of order 4, isomorphic to the Klein-4 group, and consisting of the identity permutation along with the 3 products of disjoint 2-cycles, (12)(34), (13)(24), and (14)(23).

As others have noted, any alleged subgroup H of order 6 must be isomorphic to S3, so H contains a unique subgroup T of order 3 which is thus characteristic in H. Thus T is normal in A4, being a characteristic subgroup of the normal subgroup H (of index 2) of A4.

Thus A4 is the direct product of two abelian subgroups of orders 4 and 3 and must therefore be abelian itself, which it is not. So the alleged subgroup H of order 6 does not exist.