Proposition: The group $A_4$ has no subgroup of order $6$.

Proof: Suppose we have some $H$ such that $|H|=6$, thus $[A_4 : H ] = 2$. Thus there are only two cosets of $H$ in $A_4$ . Inasmuch as one of the cosets is $H$ itself, right and left cosets must coincide; therefore, $gH=Hg$ or $gHg^{−1} = H$ for every $g ∈ A_4$. Since there are eight $3$-cycles in $A_4$, at least one $3$-cycle must be in $H$ . Without loss of generality, assume that $(123)$ is in $H$ . Then $(123)^{−1} = (132)$ must also be in $H$. Since $ghg^{−1} ∈ H$ for all $g∈A_4$ and all $h∈H$, thus $(124)(123)(124)^{−1} = (124)(123)(142) = (243)$ and $(243)(123)(243)^{−1} = (243)(123)(234) = (142)$ are in $H$. We can conclude that $H$ must have at least seven elements $(1), (123), (132), (243), (243)^{−1} = (234), (142), (142)^{−1} = (124)$. So, we come to a contradiction and $H$ can't be of order $6$.

I understand all of the proof except for "right and left cosets must coincide; therefore, $gH=Hg$ or $gHg^{−1} = H$ for every $g ∈ A_4$".

Yes, there are two distinct left cosets: ${g_1}H={g_2}H=\dots={g_k}H=H$ and ${g_{(k+1)}}H={g_{(k+2)}}H=\dots={g_{12}}H=gH$. Also, there are two distinct right cosets: $H{g_1'}=H{g_2'}=\dots=H{g_l}=H$ and $H{g_{(l+1)}}=H{g_{(l+2)}}=\dots=H{g_{12}}=Hg'$. But how we can know that

$1.\ gH=Hg'$ for the mentioned indices,

and then

$2.\ Hg'=Hg$ for the mentioned indices $?$

Please don't mark it duplicate question, since it's not about whole proof of the original Proposition.

Thank you.

  • $\begingroup$ Derek Holt; please let me know where this question has been answered, and note that my question is not about all of the theorem. $\endgroup$ – L.G. Jan 11 '15 at 11:57
  • $\begingroup$ Wouldn't this rather be duplicate of any question about "Every subgroup of index 2 is normal"? $\endgroup$ – Hagen von Eitzen Jan 11 '15 at 11:57
  • $\begingroup$ Hagen von Eitzen; the book hasn't told about normal groups yet! The proof should be based on simpler knowledge! $\endgroup$ – L.G. Jan 11 '15 at 12:00
  • $\begingroup$ I agree with Hagen von Eitzen. You have been trying to prove that all subgroups of index two are normal subgroups. The fact that you have not yet studied normal subgroups is immaterial - that is what you have been trying to prove! I am sorry if I closed the question inappropriately. I often forget that my "votes to close" result in instant closure! $\endgroup$ – Derek Holt Jan 11 '15 at 14:07
  • 1
    $\begingroup$ Derek Holt; That's alright. Thank you for your comment. The answer from Bernard is based on the fact that I already knew: "Let $H$ be a subgroup of a group $G$. Then the left(/right) cosets of $H$ in $G$ partition $G$. That is, the group $G$ is the disjoint union of the left(/right) cosets of $H$ in $G$". $\endgroup$ – L.G. Jan 11 '15 at 14:52

If $g,g'\notin H$, $gH=Hg'$ because, as the cosets constitute a partition of $G$ and there are only two cosets, both are equal to $G\setminus H$,

  • $\begingroup$ Simple Fantastic Proof! Thank you very much. $\endgroup$ – L.G. Jan 11 '15 at 12:07

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