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Let $H$ be a subgroup of the finite group $G$. For $h\in H$, let $C_h$ be the conjugacy class of $h$ in $G$. $S=\bigcup_{h\in H} C_h$. Whether $\# S\mid |G|$?

Edit after @user1729 's comment:

I encountered this question while solving this question:

Herstein ch2.12 question 12a

Let $G$ be a group of order $pqr$, $p<q<r$ primes. Prove the $r$-Sylow subgroup is normal in $G$.

My attempt to the original question:

By third part of Sylow theorem, number of $r$-Sylow subgroups is $1$ or $pq$. Suppose it is $pq$. Fix an $r$-Sylow subgroup, say $H$. Since $r$-Sylow subgroups are conjugates and since they intersect trivially,

$\sum_{i}(C_{h_i})=pqr-pq+1$, where $h_i$ is representative from a conjugacy class.

If the answer to this post is 'yes', the above equation gives a contradiction and the original question is solved.

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  • $\begingroup$ @user1729 Edited. $\endgroup$
    – Vinay Deshpande
    Jul 6, 2020 at 10:52
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    $\begingroup$ I've retracted my close vote, and turned my downvote into an upvote :-) $\endgroup$
    – user1729
    Jul 6, 2020 at 13:32
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    $\begingroup$ I am sorry but I am reluctant to retract my close vote. The problem I have is that the answer to the question is clearly yes if $G$ is abelian and, more generally, if $H$ is a normal subgroup of $G$, because in these cases we have $S=H$. So we have to look at non-normal subgroups of groups, and then the smallest example shows that the answer to the question is no. Surely, before asking a question like this, you should at least check a few small examples? $\endgroup$
    – Derek Holt
    Jul 6, 2020 at 18:22
  • $\begingroup$ @DerekHolt True. Earlier even I used to solve a problem by doing every single step myself. But this method takes a lot of time. So now, whenever I fear I'm heading the wrong way, I ask for help. In this case, while solving the original problem, I encountered the concept 'Union of conjugacy classes of a subgroup' which appeared unfamiliar, so I feared I'm on wrong path, so I didn't put much effort and asked for help. $\endgroup$
    – Vinay Deshpande
    Jul 7, 2020 at 6:37
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    $\begingroup$ OK, I have now retracted my close vote! $\endgroup$
    – Derek Holt
    Jul 7, 2020 at 7:34

1 Answer 1

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Let $G=S_3$, $H=C_2$. Then $|S|=4, |G|=6$.

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