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January 28th, 2017, 12:29 PM   #1
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Show that H is subgroup

I'm struggling with the next exercise. Can someone help me out?
Let G be a finite group. $\displaystyle \emptyset \neq H \subseteq G $ with the propriety that $\displaystyle xy \in H \quad \forall x,y \in H $. Multiplicative notation is used.
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January 28th, 2017, 01:26 PM   #2
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Since $H \neq \emptyset$ let $h \in H$ and consider $\{h, h^2, h^3, \dots\}$. Each power of $h$ is contained in $H$ by hypothesis and since $H$ is finite one of them must be the identity (why?), so $H$ is a subgroup.

For a counterexample in the infinite case take the positive integers in the group of integers with addition. The proof fails because $1 + 1 + \dots + 1$ is never $0$.

Last edited by Maschke; January 28th, 2017 at 01:35 PM.
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