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Math Help - Finite Group,

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    Finite Group,

    Let G be a finite group and let H be a non-empty subset of G that is closed under multiplication.Show that H is closed under inverse and hence prove that H is a subgroup of G.
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    Quote Originally Posted by student17989 View Post
    Let G be a finite group and let H be a non-empty subset of G that is closed under multiplication.Show that H is closed under inverse and hence prove that H is a subgroup of G.
    Clearly since |G|<\infty and H\leqslant G we have that |H|<\infty. Let h_0\in H be fixed and define \theta_{h_0}:H\to H by h\overset{\theta_{h_0}}{\longmapsto}h_0h. Clearly this is injective since \theta_{h_0}(h)=\theta_{h_0}(h')\implies h_0h=h_0h' and since G is a group we have cancellation and thus h=h'. But, every injection from a finite set to itself is a bijection. Thus, since e\in H we have that \theta_{h_0}(h)=h_0h=e for some h\in H. It follows that h=h_0^{-1} and since h_0 was arbitrary the conclusion follows.
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