# Application of Third Isomorphism Theorem: If $G'/G''$ and $G''/G'''$ are cyclic then $G''=G'''$

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.

Question

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 $$G''' = \{ e\}$$ ?

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 $$G’’’=\{e\}$$.