suppose that $G$ is a finite group and $H$ is a subgroup of $G$ that is not cyclic,I want to prove that all conjugates of $H$ are not cyclic.

Question

suppose that $G$ is a finite group and $H$ is a subgroup of $G$ that is not cyclic,I want to prove that all conjugates of $H$ are not cyclic.

I supposed that $S$ is a conjugate of $H$ and it is cyclic,I took the generator of $S$ and because for a $g \in G$ we have $S=gHg^{-1}$,so I conclude that $H$ is cyclic and we have a contradiction,I don't know my proof is right or wrong,it will be great if you guide me about this,thanks a lot.

Answer 1

What you actually proved is the contrapositive, which is still correct. It's subtle in that, you proved if a group $S$ is of the form $S=gHg^{-1}$, then $S$ being cyclic implies $H$ is cyclic.

Since our original aim was to prove, if $H$ is not cyclic, then $S=gHg^{-1}$ is not cyclic, we have in fact shown the contrapositive. You have the proof idea correct. Since $S$ is assumed cyclic, there will be some generator in $S$ of the form $x=ghg^{-1}$ and every element is of the form $x^n=gh^ng^{-1}$. Since elements of $S$ and elements of $H$ are in a one to one correspondence, this means $H$ is of the form $\langle h \rangle$.

Answer 2

Hint: For each fixed $g\in G$, show that the map $f:H\to gHg^{-1}$ given by $f(h)=ghg^{-1},\,\forall h\in H$ is an isomorphism.

Since an isomorphism preserves cyclic subgroups, the result follows (i.e. $H$ is (non-) cyclic $\Longleftrightarrow$ $gHg^{-1}$ is (non-) cyclic ).