# Alg7.5

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 $\empty \ne H\subseteq G\,$ and $\forall a,b\isin H, ab^{-1}\isin H\,$, then $H\,$ is a subgroup.

First I prove that $1_G = 1_H = 1_K\,$ which implies $H\cap K \ne \empty\,$.

$\forall a\isin G, 1_G a=a\,$ and $\forall b\isin H, 1_H b = b\,$. Since $1_H,b\isin G, 1_H b=b\implies 1_H = 1_G\,$. Similarly for $1_K\,$. Since $1_G=1_H=1_K, 1_G\isin H\cap K\ne\empty\,$.

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

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

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