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Prove that if G\, is a group and H,K\, are subgroups of G\,, then H\cap K\, is a group.

I prove it with the subgroup criteria:

If \emptyset \neq H\subseteq G\, and \forall a,b\in H,ab^{{-1}}\in H\,, then H\, is a subgroup.

First I prove that 1_{G}=1_{H}=1_{K}\, which implies H\cap K\neq \emptyset \,.

\forall a\in G,1_{G}a=a\, and \forall b\in H,1_{H}b=b\,. Since 1_{H},b\in G,1_{H}b=b\implies 1_{H}=1_{G}\,. Similarly for 1_{K}\,. Since 1_{G}=1_{H}=1_{K},1_{G}\in H\cap K\neq \emptyset \,.

Second, I prove that \forall a,b\in H\cap K,ab^{{-1}}\in H\cap K\,.

If a,b\in H\cap K\, then a,b\in H\, and a,b\in K\,. This implies ab^{{-1}}\in H\, and ab^{{-1}}\in K\, since H\, and K\, are both subgroups of G\,, so ab^{{-1}}\in H\cap K\,. Therefore, H\cap K\, is a group.

Abstract Algebra

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