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This question already has an answer here:

I'm working on a proof to show that every subgroup of an abelian group is also a normal subgroup. Let $G$ be an abelian group and $H$ an arbitrary subgroup of $G$. I want to show that $gHg^{-1} = H$, but I could use some help with some of the beginning steps to this problem. I've only recently been getting interested in Abstract Algebra.

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marked as duplicate by N. F. Taussig, Rory Daulton, Daniel Fischer Dec 24 '15 at 16:30

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First note that $gHg^{-1} = \{ghg^{-1} \mid h \in H\}$.

To show that $H$ is a normal subgroup of $G$, you need to show that, for each $g \in G$, every element of the form $ghg^{-1}$ is an element of $H$ (this shows $gHg^{-1} \subseteq H$) and every element of $H$ is of the form $ghg^{-1}$ for some $h \in H$ (this shows $H \subseteq gHg^{-1}$).

An early exercise in group theory which makes life a bit easier (which I recommend you try for yourself) is the following:

If $H \subseteq G$, then $gHg^{-1} = H$ for every $g \in G$ if and only if $gHg^{-1} \subseteq H$ for every $g\in G$.

That is, we only need to do the first of the two steps I outlined above, namely, show that every element of the form $ghg^{-1}$ is an element of $H$.

Returning to the specific case at hand, as $G$ is abelian, the elements of the group commute. So what does $ghg^{-1}$ simplify to?


Note, in this case, establishing $gHg^{-1} = H$ is no harder than showing $gHg^{-1} \subset H$, but in general it would be more difficult.

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  • $\begingroup$ If I may give an addition: this shows that $gHg^{-1} \subset H$ for all $g \in G$. This is actually sufficient to conclude $gHg^{-1} = H$. I recommend spending time to see why this is. $\endgroup$ – Eric Thoma Dec 23 '15 at 5:16
  • $\begingroup$ @EricThoma: In general that's true, but in this case (as was pointed out to me by rschwieb), it gives $gHg^{-1} = H$ almost immediately. $\endgroup$ – Michael Albanese Dec 23 '15 at 5:17
  • $\begingroup$ I agree. I would recommend working this out in the general (non-commutative) case, as it may help the questioner understand more about conjugation. But I concede my suggestion is confusing in this particular case. $\endgroup$ – Eric Thoma Dec 23 '15 at 5:20
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    $\begingroup$ @EricThoma: I have edited my answer to make the situation more clear. $\endgroup$ – Michael Albanese Dec 23 '15 at 5:47

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