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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.


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.

  • $\begingroup$ 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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