There are two statements that I have to determine if they are true or false:

The commutator subgroup of a simple group $G$ must be $G$ itself.


The commutator subgroup of a nonabelian simple group $G$ must be $G$ itself.

Here's my attempt to answer:

For the second, since $G$ is simple, the only factor group is $G/G\simeq \lbrace e \rbrace$, which is abelian and therefore the commutator group is contained in the normal group $G$. Now, every $x\in G$ fulfills $xg=gx$ and therefore is of the form $xgx^{-1}g^{-1}=e$, which is in $C$. Thus $C=G$.

Now here is my question. The first statement comprises abelian and nonabelian groups. In the second, we have seen that $C=G$. In order for the first statement to be true, we need to check only that abelian groups have as commutator group again $G$. Why is this not the case?


The first statement is simply false for non-trivial abelian simple groups $G$, because then $[G,G]$ is trivial. For the second statement, $[G,G]$ is a non-trivial normal subgroup in $G$, hence equal to $G$, since $G$ is simple.


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