Question
Let $G$ be a group of order 100 that has a subgroup $H$ of order 25 . Prove that every element of $G$ of order 5 is in $H$.
Step 1
Let $n_5$ denote the number of Sylow 5-subgroups of $G$. By Sylow's Theorem, we know that $n_5$ divides the order of $G$, and $n_5 \equiv 1 \pmod{5}$. Since the order of $G$ is 100, the only possibilities for $n_5$ are 1 and 4. Show more…
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Let $G$ be an abelian group. If $H=\left\{x \in G: x=x^{-1}\right\}$, that is, $H$ consists of all the elements of $G$ which are their own inverses, prove that $H$ is a subgroup of $G$.
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