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This question is related to this other post: I was wondering if proving that $ H \trianglelefteq G$ and $H\cap G^{\prime} =\{e\}$ (where $G^{\prime}$ denotes the commutator subgroup of $G$) implies that the elements of $H$ commute with the elements of $G^{\prime}$ is the same as proving that $H \trianglelefteq G$ and $H \cap G^{\prime}=\{e\}$ imply that $H \subseteq C(G)$ (where $C(G)$ is the center of $G$)? If not, is there a way to modify/extend this proof to show that $H \subseteq C(G)$? (It just needs to be a subset, not a subgroup).

If still not, how can I prove that$H \trianglelefteq G$ and $H \cap G^{\prime}=\{e\}$ imply that $H\subseteq C(G)$?

Thank you.

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  • If $H$ is a subgroup of $G$ and a subset of the center $Z(G)$ which itself a subgroup of $G$, that automatically means $H$ would be a subgroup of $Z(G)$, so it's a bit jarring to see you say "it just needs to be a subset, not a subgroup." Also, "elements of $H$ commute with elements of $G',$" is equivalent to $H\subseteq C(G')$ which is generally a weaker condition than $H\subseteq Z(G)$, so more work will have to be done to show $H$ is central (if that does indeed follow from the hypotheses). – anon Jan 02 '17 at 05:37
  • @arctictern When you say $C(G^{\prime})$ you mean $Z(G^{\prime})$ by your notation? –  Jan 02 '17 at 05:46
  • No. $C(G')$ is short for $C_G(G')$ which means the set of all $g\in G$ which commute with every element of $G'$. The notation $Z(G)$ means $C_G(G)$, the center, containing all $g\in G$ which commute with every element of $G$. Thus, if someone wrote $Z(G')$, they would be talking about the elements of $G'$ which commute with every element of $G'$, and that is not a thing being discussed. – anon Jan 02 '17 at 05:52

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Suppose $h\in H$ and $g\in G$ are arbitrary, and consider the commutator $[g,h]=(ghg^{-1})h^{-1}$.

Using the fact $H$ is normal, can you deduce anything? (Hint: look at how I parenthesized it.)

anon
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    what you parenthesized is contained in $H$ since $H$ is normal? And so $[g,h]$ is contained in $H$? –  Jan 02 '17 at 05:47
  • I made an edit to that comment. –  Jan 02 '17 at 05:51
  • Yes. As $H$ is normal, $ghg^{-1}\in H$, so $[g,h]=ghg^{-1}h^{-1}\in H$. Now what can you deduce from here? – anon Jan 02 '17 at 05:52
  • the commutator of $G$ and $H$ Is contained in $H$? –  Jan 02 '17 at 05:55
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    You already deduced $[g,h]$ is in $H$. Go further. Note that $[g,h]$ is a commutator. Is there anything in our set of hypotheses that talks about commutators and $H$? – anon Jan 02 '17 at 05:57
  • they intersect trivially which seems kind of contradictory.., –  Jan 02 '17 at 05:58
  • Not contradictory. Just restrictive. What can we say now about $[g,h]$? – anon Jan 02 '17 at 05:58
  • that it is the identity? –  Jan 02 '17 at 05:59
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    Yes. If $[g,h]$ is in $G'$ (which it is since it's a commutator) and $[g,h]$ is in $H$, and the only thing in both $G'$ and $H$ is the identity, then $[g,h]$ is the identity. What does that say about $g$ and $h$? – anon Jan 02 '17 at 06:00
  • two elements $x$ and $y$ of a group $G$ commute iff $[x,y]=e$. Heyy! –  Jan 02 '17 at 06:04
  • So, $H \subseteq C(G)$... –  Jan 02 '17 at 06:05
  • Yes, $[g,h]=e$, or in other words $ghg^{-1}h^{-1}=e$, is the same as $gh=hg$, i.e. $g$ and $h$ commute. Since $g\in G$ and $h\in H$ were arbitrary, that means every element of $H$ commutes with every element of $G$, so $H$ is contained in the center of $G$ (which I and most people denote $Z(G)$.) – anon Jan 02 '17 at 06:06
  • you, most people, but not my professor. Thank you!! –  Jan 02 '17 at 06:07