Exercise 4.5.26 (If $|G| = 105$, and if a Sylow $3$-subgroup of $G$ is normal, then $G$ is abelian.)

Question

Let $G$ be a group of order $105$. Prove that if a Sylow $3$-subgroup of $G$ is normal, then $G$ is abelian.

Richard Ganaye · Accepted Answer

\begin{proof} 
Here $|G| = 3 \cdot 5 \cdot 7$. 

By hypothesis, $G$ has a unique Sylow $3$-subgroup $P$, of order $3$, which is normal in $G$.

Since $P \unlhd G$, $G$ acts on $P$ by conjugation. This action affords an homomorphism
$$
\varphi 
\left\{
\begin{array}{ccl}
 G &	o &\mathrm{Aut}(P)\
 g & \mapsto &\varphi_g
      \left\{
      \begin{array}{ccl}
      P & 	o &P\
      x & \mapsto & gxg^{-1}
      \end{array}
      ight.
 \end{array}
 ight.
 $$

\bigskip
Since $P \simeq Z_3$, 
$$|\mathrm{Aut}(P)| = 2.$$
Then $|G| = 105$ is relatively prime to $|\mathrm{Aut}(P)| = 2$, hence the homomorphism $\varphi$ is trivial.

Then for all $g \in G$, and for all $x \in P$, $gxg^{-1} = x$. This shows that
$$P \leq Z(G).$$

\bigskip
Put $\overline{G} = G/P$. Then $|\overline{G}| = 5\cdot 7 = 35$. 

By the first example (``groups of order pq'', p. 143), since $p = 5 
mid q-1 = 6$, $\overline{G}$ is cyclic, so 
$$\overline{G} = G/P \simeq Z_{35}.$$

By the Third Isomorphism Theorem, since $P \leq Z(G)$,
$$ (G / P) /(Z(G)/P) \simeq G/Z(G).$$
So $G/Z(G)$ is a quotient group of the cyclic group $G / P$, so is cyclic. By Exercise 3.1.36, this proves that $G$ is abelian.
\end{proof}

Exercise 4.5.26 (If $|G| = 105$, and if a Sylow $3$-subgroup of $G$ is normal, then $G$ is abelian.)

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Exercise 4.5.26 (If $|G| = 105$, and if a Sylow $3$-subgroup of $G$ is normal, then $G$ is abelian.)

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