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Seeking contradiction assume that $A_4$ has a subgroup of order $6$. Clearly all elements of $A_4$ have order $1$, $2$ or $3$. Thus the subgroup of order $6$ cannot be cyclic. Thus it must be isomorphic to $S_3$. In particular it must have two elements of order $3$ and three elements of order $2$.

I show that if $h$ is any element of $A_4$ of order $3$ and $g$ any element of $A_4$ of order $2$, then $hg$ is of order $3$.

First $hg$ cannot be of order $1$. Since that would mean that $h = g^{-1}$. And since the order of the inverse is the same as that of the element, this would imply that $h$ and $g$ have the same order.

Further, $hg$ cannot have order $2$. Let's assume, for the sake of contradiction that it does. Let's name $h_1$, $h_2$ and $h_3$ any distinct elements of $A_4$ of order $3$. And $g$ any one of the three elements of $A_4$ of order $2$. Then $h_1g$, $h_2g$ and $h_3g$ are $3$ distinct elements such that $h_1g \neq g$ and $h_2g \neq g$ and $h_3g \neq g.$ By our assumption each of $h_1g$, $h_2g$ and $h_3g$ have order $2$. Since none of them equal to $g$ we have four elements of order $2$. But $A_4$ has only three elements of order $2$. This is a contradiction. Also, $hg$ cannot have order greater than $3$ since no element of $A_4$ has order greater than $3$.

Let's give the elements of our subgroup names: $e$, $m_1$, $m_2$ and $k_1$, $k_2$, $k_3$. The m's are the two elements of order $3$ and k's are the three elements of order $2$.
Finally $m_1k_1$, $m_1k_2$ and $m_1k_3$ are three distinct elements of our subgroup of order $3$. But this is a contradiction with the fact that the subgroup has only two elements of order $3$. Thus $A_4$ cannot have a subgroup of order $6$.

Adam
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  • Did you already compare it yourself with the standard proofs at this site, e.g. here, or here, etc.? – Dietrich Burde Sep 28 '18 at 13:22
  • Yep, but I do not understand cosets and normal subgroups so they may be the same without me knowing it. – Adam Sep 28 '18 at 13:23
  • There are many answers not using cosets or normal subgroups, too. On the other hand, you probably should understand normal subgroups before paying attention to this question here. What about the answer of azimut? – Dietrich Burde Sep 28 '18 at 13:25
  • I already knew about azimut's way, it is the answer in my textbook. And while my proof is similar in that it also uses isomorphism with $S_3$, I believe it is a different proof. – Adam Sep 28 '18 at 13:50

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An easier proof along the same lines is to note that no two elements of order two in $S_3$ commute, lest they generate a subgroup of order 4, which does not divide 6. Now $S_3$ and $A_4$ each have 3 elements of order 2, so pick any two elements of order two in $A_4$, note that they commute, and you are done.

C Monsour
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