1
$\begingroup$

When asked to show that a subgroup $H$ of the abelian group $G$ is normal, does it suffice to say:

  • first, $H$ is a subgroup, so it contains the identity element of $G$ and inverses $h^\prime$ for every element $h$ in $H$;
  • because $G$ (and thus $H$) is abelian, every pair of elements of $H$ commute with each other. Then every element in $H$ commutes with any element of $G$, as the elements of both $H$ and $G$ all commute with any other element of $G$ (or $H$).

Thus $gH = Hg$ and $H$ is a normal subgroup.

I suspect the above is insufficient. In particular, any potential errors above aside, I was wondering if it is "necessary" to show that $ghg' = h$?

$\endgroup$
2
  • $\begingroup$ Your instincts are sound. I'd rephrase your second point slightly; as in "because G is abelian, every element of G commutes with every element of H, hence conjugation acts trivially on H" but I think you got the point across. $\endgroup$
    – lulu
    Jul 6, 2015 at 0:46
  • $\begingroup$ You don't need to show that $ghg^{-1}=h'$ because if this is true for an arbitrary $h$, it means that $gHg^{-1}=H$, or, equivalently, that $gH=Hg$, which you showed. In general, $gH=Hg$ for all $ g \in G$ if and only if for all $h \in H$ and $g \in G$ $ghg^{-1}=h'$ for some $h' \in H$. These are just two ways to show $H$ is normal in $G$. $\endgroup$
    – coldnumber
    Jul 6, 2015 at 1:28

3 Answers 3

1
$\begingroup$

It's a good attempt, but your argument is unnecessarily complex and seems to me to use some incorrect logic:

  • there's no need to use the fact that $H$ is a subgroup, since $gS=Sg$ for any $g\in G$ and any subset $S\subset G$ whatsoever when $G$ is abelian

  • you seem to be saying the fact that every element of $H$ commutes with every other element of $H$ implies that every element of $H$ commutes with every element of $G$, when in fact that is not a correct implication (moreover, both are subsumed by the simple statement "$G$ is abelian")

(Of course, these criticisms of the argument not affecting the fact that the result is true.)

Let's simplify:

For any $g\in G$, we have $$gH=\{gh:h\in H\}\overset{\text{(because $G$ is abelian)}\strut}=\{hg:h\in H\}=Hg$$ and therefore $H$ is normal.

$\endgroup$
0
$\begingroup$

Variant: In an abelian group, conjugation is trivial (= identity), hence any subgroup is stable by conjugation, i.e. is normal.

$\endgroup$
0
$\begingroup$

Your argument is fine, if a bit wordy. The Reader's Digest version would suffice:

$G$ abelian $\implies gh = hg$ for all $h \in H$, and any $g \in G \implies gH = Hg$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .