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I tried using the result that if p is the least prime dividing the order of G, then any subgroup of index p is normal in G. So a subgroup H of order 3 is normal in G. Then, if there was a normal subgroup K of order 2 then G=HK. But how do I get a contradiction?

Ishan Srivastava
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  • There is only one non-abelian group of order $6$, up to isomorphism. No need to invoke Sylow or lowest prime dividing the order or any such stuff here, you just check. – Arthur Oct 09 '17 at 13:29
  • Do you know Cauchy's theorem, which says that for every prime $p$ dividing the order of $G$, there must be in $G$ an element of order $p$? – Arnaud D. Oct 09 '17 at 13:41
  • Yes I know. So how should I use it here? – Ishan Srivastava Oct 09 '17 at 13:44

2 Answers2

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Consider the conjugacy class of an element of order $2$. If there is more than one element in it, the subgroups generated by each element of the conjugacy class have order $2$ and are conjugate to each other - hence not normal.

If, on the other hand, the conjugacy class has just one element in, then the element is in the centre of the group and commutes with an element of order $3$. These two elements therefore generate a cyclic group of order $6$ which must be the whole group and is abelian.

[Alternatively, once you know the centre is non-trivial you know that $G/Z$ must be cyclic and hence $G$ must be abelian].

Mark Bennet
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By Cauchy's theorem, there must be an element of order $2$, and thus a subgroup $K$ of order $2$. Then your argument shows that it cannot be normal, because $HK=G$ would be abelian.

Arnaud D.
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    Okay got it. Thanks! – Ishan Srivastava Oct 09 '17 at 15:36
  • @IshanSrivastava Since this is your first question, I would like to suggest you to read this page of the FAQ : What should I do when someone answers my question? – Arnaud D. Oct 09 '17 at 15:46
  • @ArnaudD. I don't see why $G=HK$ implies $G$ is abelian. Why is that true? – user193319 Jan 30 '18 at 14:37
  • @user193319 It doesn't, in general; the point is that if $H$ and $K$ are normal and $H\cap K$ is trivial (as is the case here), then $HK\cong H\times K$. Since in this case $H$ and $K$ are cyclic their product would be abelian. I should have probably written this explicitly in my answer, though... – Arnaud D. Jan 30 '18 at 14:46
  • @ArnaudD. Ah! I see. There's one thing that still confuses me, however: it seems that I can prove $G$ is abelian regardless of whether $K$ is normal. Let $H={1,y,y^2}$ and $K={1,x}$. Since $H$ has an index of $2$, $H$ is normal and therefore $KH$ is a subgroup. Suppose that $a \in H \cap K \setminus {1}$. Then $a=x$ and $a=y$ or $a=y^2$. Since $x\neq y$, $a=y^2$. But this implies $2=|x|=|y^2|$ and therefore $e = (y^2)^2 = y$, a contradiction. Hence $H \cap K = {1}$, which implies $KH$ is a subgroup of order $6$. So $G = KH \simeq K \times H$ is abelian...Where'd I go wrong? – user193319 Jan 30 '18 at 16:39
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    If $K$ is not normal, you don't have an isomorphism with $K\times H$ but rather with $K\ltimes H$. This should appear on some question here. – Arnaud D. Jan 30 '18 at 19:11