Question
Let $|G|=p^{n} m$, where $p$ is prime and $\operatorname{gcd}(p, m)=1$. Suppose that $H$ is a normal subgroup of $G$ of order $p^{n} .$ If $K$ is a subgroup of $G$ of order $p^{k}$, show that $K \subseteq H$.
Step 1
Since $H$ is a normal subgroup of $G$ of order $p^n$, we know that $H \trianglelefteq G$ and $|H| = p^n$. Show more…
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Let $G$ be an abelian group. Let $H$ be a subgroup of $G$, and let $K$ consist of all the elements $x$ in $G$ such that some power of $x$ is in $H$. That is, $K=\left\{x \in G:\right.$ for some integer $\left.n>0, x^{n} \in H\right\}$. Prove that $K$ is a subgroup of $G$.
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