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Suppose $G$ is a group. H is a subgroup of $G$. Now determine whether $\cup_{x\in G} xHx^{-1}$ is a subgroup.

I tend to proof this is not true since I know the union of any two subgroups which do not contained in another is not a subgroup. But I don't know if it is true for countable unions. I need a counterexample such that $\cup_{x\in G} xHx^{-1}$ is not a subgroup

Jaqen Chou
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    This could be a subgroup, for examples if $H$ is the trivial subgroup of $G$, then the union is always a subgroup. The question is not completly clear: Should you proof the statement or find a counterexample? Or should you find a criterion when the statement holds? – Babelfish Oct 01 '18 at 13:06
  • This might partially answer your question. https://math.stackexchange.com/q/75613 – JB071098 Oct 01 '18 at 13:07

1 Answers1

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If $H$ is not a normal subgroup, we have a nontrivial union of subgroups, which isn't necessarily a subgroup.

For the smallest example, take $G:=S_3$ and $H:=\langle (1\,2)\rangle =\{id, \ (1\,2)\}$. Its conjugates are $$\{id,\ (2\,3)\},\ \ \{id,\ (3\,1)\}$$ Their union consists of $4$ elements, and anyway is clearly not closed under multiplication.

Berci
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