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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

Here is a proof by class equation.

Notice that the class equation for $A_{4}$ is $|A_{4}| = 12 = 4+4+3+1$. As the OP pointed out, if $H<A_{4}$ has order 6 then $H$ is a normal subgroup. Therefore, for every $\sigma\in H$, all elements in $\sigma$'s conjugancy class would lie in $H$. This means that 6 can be partitioned by the RHS of our class equation, which is clearly impossible.

For how we derive the class equation for $A_{4}$, we first get one for $S_{4}$ by writing all possible cycle types and decide whether there exists some odd permutation $\tau$ such that $\tau^{-1}\sigma\tau = \sigma$, which tells us if that conjugancy class "breaks" into two or remains as one in $A_{4}$. For this particular example, we do not even need to try to find any $\tau$ and $\sigma$ (write everything out for $S_{4}$ would suffice)

Answer 6

Here's another approach.

We denote $G=A_4$ and $V_4$ the Klein four-group (which is a normal Sylow $2$-subgroup of $G$).

Suppose that there exists $N\leq G$ such that $|N|=6$. Then $|G:N|=2$ and $N\unlhd G$. If $Q\in\text{Syl}_3(G)$ then $Q\in\text{Syl}_3(N)$, as $N\unlhd G$. By Frattini's argument $G=NN_G(Q)$. If $|N_G(Q)|=12$ then $Q\unlhd G$ and $G\cong V_4\times Q$, but this is not possible as $A_4$ is not abelian. If $|N_G(Q)|=6$ then $N_G(Q)\unlhd G$ and

$$|N\cap N_G(Q)|=\frac{|N||N_G(Q)|}{|NN_G(Q)|}=\frac{36}{12}=3.$$ Thus $N\cap N_G(Q)\in\text{Syl}_3(G)$ is normal in $G$ and $G\cong V_4\times(N\cap N_G(Q))$, which is not possible as $A_4$ is not abelian. Finally, if $|N_G(Q)|=3$ then

$$|N\cap N_G(Q)|=\frac{|N||N_G(Q)|}{|NN_G(Q)|}=\frac{18}{12}=3/2,$$ which is a contradiction.

Answer 7

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.

Answer 8

This wasn't straightforward for me, and so I wrote a more detailed solution in hopes of helping novice students like myself. Please correct me if I did something wrong, or ended up complicating things.

Assume that there exists $H \leq A_4$ such that $|H| = 6$. Given the formula: $|A_n| = \frac{|S_n|}{2} = \frac{n!}{2}$ it follows that: $|A_4| = 12$

From Lagrange's theorem we get that: $|A_4 \mathbin{/} H|=\frac{|A_4|}{|H|}=2$

Thus: $H\trianglelefteq A_4$ and $A_4 \mathbin{/} H$ is well defined.

$|H| = 6$, and we know that: $(1 2 3), (1 3 2), (1 2 4), (1 4 2), (2 3 4), (2 4 3), (1 2 4), (1 4 2) \in A_4$. In total these were $8$ cycles of size $3$ in $A_4$. From this we gather that there exists $\pi \in A_4 \setminus H$ such that $\pi$ is of length 3.

We know that $\pi H \neq H$ and so $\pi^2H \neq \pi H$ and so $\pi^2H=H$ and at last $\pi^3H=\pi H$ however $\pi^3=e$ in contradiction to $\pi H \neq H$ because given $\pi = (a b c)$ we get that $\pi^2 = (a c b)$ and that $\pi^3 = \pi \cdot \pi^2 = (a b c)(a c b) = (a)(b)(c) = e$.

At last $\pi^3H = eH$

Answer 9

If possible, let there exists a subgroup $H$ of $A_4$ with $|A_4:H|=2$. Then $H\triangleleft G$ and $|G/H|=2$. By Lagrange's theorem $g^2\in H$ for all $g\in A_4$. Thus for any $3$-cycle $(a\ b\ c)\in A_4$, $(a\ b\ c)=(a\ c\ b)^2\in H$. Therefore all $3$-cycles being in $A_4$ must also be in $H$. However, the number of $3$-cycles in $A_4$ equals to ${4 \choose 3}\times 2!=8$. This shows that $H$ has at least $8$ elements, contradicting that $|H|=\frac{12}{2}=6$. Thus H does not exist.

Answer 10

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.

Answer 11

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.