Show a proof that $Z(G),$ the center of $G,$ is a normal subgroup of $G$ and every subgroup of the center is normal. I know that $Z(G)$ is a normal subgroup and it is abelian, but how would I show that every subgroup of $Z(G)$ is also normal?
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1Igor's answer is great. I only want to mention that you can get the same result by powerful machinery that you will probably learn in your future, which has much wider application. That is the following: If $H$ is normal in $N$, and $N$ is characteristic in $G$, then $H$ is normal in $G$. The center $Z(G)$ is characteristic in $G$, but so are many other subgroups of $G$. – Doc Dec 03 '13 at 17:27
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@Doc hi, It has been 12 years, but I believe that is false, no? It is true that if $H$ is characteristic in $N$, and $N$ is normal in $G$, then $H$ is normal in $G$. But not what you wrote (I believe https://math.stackexchange.com/questions/4252458/normal-subgroup-of-a-characteristic-subgroup provides a counterexample). – Neckverse Herdman Dec 15 '24 at 10:57
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Since if $H<Z(G),$ and $h \in H,$ for every $g \in G$ $g^{-1}hg = h,$ (since $h$ is central), it follows that $g^{-1}H g = H$ (it is true elementwise).
Igor Rivin
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Recall that the center of a group $Z(G)$ consists of all elements that commute with every element in $G$. That's pretty much all you need, here:
If $H<Z(G),$ and $h \in H,$ then $h \in Z(G),$ and so for every $g \in G,\;$ $g^{-1}hg =g^{-1}gh = eh = h$ ($h \in Z(B)$, so $h$ commutes with $g$.)
Since this is true of all $h \in H$, we have that $g^{-1}Hg = H$, and hence, $H$ is normal.
amWhy
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1@Prahlad. Ultimately, yes. But I just wanted to explicitly state why $g^{-1}hg = h$, using the definition of what it means to be an element in the center of the a group: that it commutes with every element in the group. – amWhy Dec 03 '13 at 17:35
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