If $H$ is the unique subgroup of its size, $H Actually from the above condition I will conclude that $H \triangle G$. So I cannot use that since $H \triangle G$ $\implies gH = Hg$ $\implies gHg^{-1} = H$. Any ideas?
Unique subgroup $H$ of $G$ $\implies gHg^{-1} = H $ $\forall g \in G$.
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abstract-algebra
group-theory
finite-groups
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0Do you mean $H \lhd G$? It's produced by `$H \lhd G$` although I'm not sure what it stands for (maybe left hand diamond?). – 2017-01-28
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0I think both means the same [here] , that is $H$ is a normal subgroup of $G$. Is it not? otherwise i will edit it! – 2017-01-28
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0I've never seen an upright triangle used to denote normality (maybe it is used by some, now I'm curious!) I've only ever seen the triangle "pointing at" the normal subgroup. – 2017-01-28
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0If you look at Wikipedia article [Normal subgroup](https://en.wikipedia.org/wiki/Normal_subgroup), it indeed uses notation $H\lhd G$. Anyway, whichever notation you use, you should probably explain in the question what it means. (Especially in the case that the notation is non-standard, which seems to be the case here.) – 2017-02-04
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0Somewhat related older post: [If G is a finite group, and $\exists !$ $H\leq G$ such that $|H| =n$, then $H \triangleleft G$.](http://math.stackexchange.com/q/361300) – 2017-02-04
1 Answers
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Let $K=gHg^{-1}$. Then $K$ is a subgroup of $G$, and the map $H\to K$ given by $h\mapsto ghg^{-1}$ is a bijection since it has inverse $k\mapsto g^{-1}kg$.
Therefore $|K|=|H|$, and since $H$ is the unique subgroup of $G$ having cardinality $|H|$ it follows that $K=H$.