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Let $G$ be a group and let $H$ be a normal subgroup of order 2. Prove that every element of $H$ commutes with every element of $G$.

Since $H$ is a normal subgroup of order 2, $H$ contains two elements, one of which is an identity element (as $H$ is, of course, a group). So $H = \{e,h\}$, for some other non-identity element $h \in H$. And $H$ is normal, so

\begin{align} Hg = gH &\implies \{e,h\}g = g\{e,h\} \\ &\implies \{g,hg\} = \{g,gh\} \end{align} for some $g \in G$.

Does this mean we possibly have $g=g$ (trivially), $g = gh$, $hg = g$, and $hg = gh$, and I need to prove each one of them? Furthermore, with $hg = gh$, is it necessary to show that $H$ is a central subgroup as well?

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Alternatively, an easier condition for normality to work with here is $ghg^{-1}\in H$ for all $g\in G$ and all $h\in H$. Note that if $ghg^{-1} = e$ then $h=e$. –  Tobias Kildetoft Feb 19 at 7:55

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Continuing from your work- If $\{g,hg\}=\{g,gh\}$ then you can deduce that we must have $gh=hg$ (since we know we already have $g=g$, and both sets must be equal since $H$ is a normal subgroup). This gives you already that the only nontrivial element of $H$ commutes with all elements of $G$.

Since, the only other element of $H$ is the identity element $e$, and this commutes with everything in the group trivially, we are done.

There is no need to go through all the equalities $g=gh,g=g,hg=g,hg=gh$. Since we know we already have $g=g$, this guarantees $hg=gh$ for every $g\in G$. And in deducing this you have already shown that $H$ "is central" by definition. Indeed, since $h$ commutes with every element of $G$ it belongs to the center of $G$, and so does $e$ (trivially), so $H$ is contained in the center.

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