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 January 28th, 2017, 12:29 PM #1 Newbie   Joined: Jan 2017 From: Nowhere Posts: 5 Thanks: 1 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.
 January 28th, 2017, 01:26 PM #2 Senior Member   Joined: Aug 2012 Posts: 2,305 Thanks: 705 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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