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If $(G,\oplus)$ is a group, let $H=\{a\in G\mid g\in G:a\oplus g=g\oplus a\}$. Prove that $(H,\oplus)$ is a subgroup of $(G,\oplus)$.

Isn't $H$ just $G$ with a different name. In other words, does it have the same members?

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    $\begingroup$ If $G$ is abelian, yes. If it isn't abelian, then no. As an example, let $G$ be the group of all invertible $2\times 2$ matrices with real entries. Is $H = G$ in that case? $\endgroup$ Dec 14, 2016 at 3:03
  • $\begingroup$ No. $G$ is not necessarily commutative. Think about $G=S_3$ - what does $H$ look like in that case? $\endgroup$ Dec 14, 2016 at 3:03

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First, note the very useful comments made on your post. In this case, $H$ consists of the elements that commute with everything in $G$, and is known as the center of $G$, often denoted $Z(G)$. Here is the proof:

Take some $g, h \in H$. Then for any $x \in G$, $$(g \oplus h) \oplus x = g \oplus h \oplus x = g \oplus x \oplus h = x \oplus (g \oplus h),$$ so $g \oplus h \in H$.

Now see if you can prove also that $g^{-1}$ is in $H$, and you are done.

If you get stuck, see this post that has the exact same question.

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  • $\begingroup$ Okay also sorry i just saw that you're supposed to avoid saying stuff like thanks in the comments. Last question, why would commutativity and associativity imply that gh is a member of H? I just started studying groups and may have missed something crucial. I could just take your word for it but would like to understand. $\endgroup$
    – Will
    Dec 14, 2016 at 4:00
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    $\begingroup$ the fact that $(g \oplus h) \oplus x = x \oplus (g \oplus h)$ shows that $g \oplus h$ meets the conditions for being in the set $H$, by the definition of $H$. $\endgroup$ Dec 14, 2016 at 4:07

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