Let $G$ be a group with $|G|=60=2^2 \cdot 3 \cdot 5$ and assume $G$ has a unique $3$-Sylow subgroup $H_3$. The claim is that then $G$ has also a unique $5$-Sylow subgroup.
This is the procedure to prove it:
i) Show that $H_3$ is normal in $G$. (Follows directly since all Sylow subgroups are conjugates).
ii) Show that $G/H_3$ has a unique $5$-Sylow $H_5$ (Follows from $|Syl_5(G)|\equiv 1 \, mod \, 5$ and $|Syl_5(G)|$ divides $4$).
Let $H$ be the inverse image of $H_5$ by the projection $ \pi :G\longrightarrow G/H_3$. Now the goal is to note how many Sylow-$5$ subgroups there are in $H$ and use this to prove the claim. I don't understand this last step at all.
Thank you for any clarifications.
Best Answer
Since $H_5$ is normal in $G/H_3$ the subgroup $H$ is normal in $G$. Furthermore $H$ has order $15$ and therefore has a unique Sylow $5$-subgroup $U$. Let $V$ be any Sylow $5$-subgroup of $G$. Since all Sylow $p$-subgroups are conjugate to each other there is some $g\in G$ with $V=gUg^{-1}$. Since $H$ is normal in $G$ we get $V\leq H$, hence $V$ is a Sylow $5$-subgroup of $H$ and therefore $V=U$, so $G$ has a unique Sylow $5$-subgroup.