Dummit and Foote, 6.1.18

Suppose both $G'/G''$ and $G''/G'''$ are cyclic, then $G''=G'''$ (you may assume $G'''=1$).

Where $G' = [G,G], G'' = [G',G'],$ etc.


The assumption that $G'''=1$ apparently follows from the third isomorphism theorem, but I don't see how.

I know $G', G''$, and $G'''$ are all characteristic (and hence normal) in $G$.

Then the third isomorphism theorem says $$ G''/G''' \triangleleft G/G''' $$ and $$ (G/G''') \; \big{/} (G'/G''') \cong G/G' $$ I don't see why that this implies $G''' = \{ e\}$ ?


1 Answer 1


It’s not that the assumptions imply that $G’’’=\{e\}$. Instead, it’s that if you can prove it in the case that $G’’’=\{e\}$, then the result will hold in all cases, not just those in which $G’’’=\{e\}$.

To verify this, assume that you know the result holds if $G’’’=\{e\}$, and let $K$ be a group such that $K/K’$ and $K’/K’’$ are both cyclic. We want to show that $K’’=K’’’$.

Let $G=K/K’’’$. Then $G’ = (K/K’’’)’ = K’/K’’’$ and hence $G/G’ = (K/K’’’)/(K’/K’’’) \cong K/K’$ is cyclic, and $G’’=(K/K’’’)’ = K’’/K’’’$, so $G’/G’’ = (K’/K’’’)/(K’’/K’’’) \cong (K’/K’’)$ is cyclic. Thus, $G$ satisfies the hypotheses of the result and $G’’’=K’’’/K’’’$ is trivial, so given our assumption that the result holds when $G’’’$ is trivial, we conclude that $G’’=G’’’$. But this means that $K’’/K’’’ = G’’$ is trivial, so $K’’=K’’’$... which is what we wanted to show.

Thus, if you can prove the result when $G’’’=\{e\}$, then the result always holds. Thus, we may assume that $G’’’=\{e\}$.


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