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I could not solve it properly: $\bigcup_{x \in G} xHx^{-1} \neq G$ if $G$ is a finite group and $H$ is a proper subgroup (of $G$).

I tried to use the class equation and to create other actions to somehow prove this inequality, but nothing worked until now.

Thanks in advance.

user40276
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  • $\bigcup_{x \in G} xHx^{-1}={xhx^{-1}|x\in G, h\in H}\subset G $ but there are elements $g$ in $G$ that need not be written $g=xh x^{-1}$ for some $x\in G$ and $h\in H$. For example in $G=S_3$ the element $(12)$ is not conjugate to any element of $H=\mathbb Z_3=<(123)>$. – palio Apr 30 '13 at 18:49
  • This is a duplicate of this question http://math.stackexchange.com/questions/374078/reference-of-a-theorem-in-group-theory – palio May 01 '13 at 07:20

1 Answers1

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The subgroup $H$ has $[G : N_G(H)]$ distinct conjugates. Each $xHx^{-1}$ has order $|H|$.

You can use these two facts to find an upper bound for the number of elements in $\bigcup_{x \in G} xHx^{-1}$.

Mikko Korhonen
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