If $G$ is a group of order $5\cdot11\cdot13^2$ with an element of order 55 then $G$ is abelian I'm doing some exercises by myself to prepare for an exam, could you help me to prove that if $G$ is a group of order $5\cdot11\cdot13^2$ with an element of order 55 then $G$ is abelian, please?
 A: The following approach is pretty artless, but it does have the advantage of being direct. Keep Prof Emerton's comments in mind: finite abelian groups are products of various $\mathbf Z/p^n\mathbf Z$, where $p$ is a prime number.
There's a cyclic subgroup $H$ of order $55$. Consider a $13$-Sylow subgroup $P$. The number of such subgroups is $\equiv 1 \bmod 13$ and equals the index $(G : N(P))$, where $N(P)$ is the normalizer of $P$ in $G$. Use these facts to show that $P$ is normal.
It follows that $G$ is a semidirect product $P \rtimes H$. This is simply the direct product $P \times H$ if the corresponding homomorphism $H \to \operatorname{Aut}(P)$ is trivial. Can we show that this is forced upon us by the numbers? You can show that $P$ is isomorphic to either $\mathbf Z/13^2\mathbf Z$ or $(\mathbf Z/13\mathbf Z)^2$. If $P$ is cyclic then $\operatorname{Aut}(P) \approx (\mathbf Z/13^2\mathbf Z)^*$ has order $\varphi(13^2) = 13(13 - 1)$; otherwise, it is the general linear group $GL_2(\mathbf Z/13\mathbf Z)$, whose order is known.
A: You could also proceed as follows: The number of Sylow $p$-subgroups is a divisor of $169$ for $p = 5$ and $p =11.$ Hence, by Sylow's theorem, there is one Sylow $5$-subgroup and one Sylow $11$-subgroup. The number of Sylow $13$ subgroups is a divisor of $55,$ so must again be $1$. Hence $G$ is the direct product of its Sylow subgroups. Since each Sylow subgroup of $G$ is Abelian, $G$ itself is Abelian.  
A: The number of Sylow $11$-subgroups divides $5 \cdot 13^2$ and is $\equiv 1 \mod 11$. Therefore there is only one Sylow $11$-subgroup, which has to be normal because Sylow subgroups of the same order are conjugate. By the same argument the Sylow $13$-subgroup is normal. 
So there exist $H, N \trianglelefteq G$ with $|H| = 11$ and $|N| = 13^2$. You can show that any group of order $5 \cdot 13^2$ is abelian, and thus $G/H$ is abelian. This shows that $G' \leq H$, where $G'$ is the commutator subgroup. 
Therefore $G' \cap N = \{1\}$. Since $N$ is also normal, $N$ must be central: for every $g \in G$ and $n \in N$ we have $g^{-1}n^{-1}gn \in G' \cap N$ and so $gn = ng$. 
Finally we use our assumption: there exists an abelian subgroup $M$ of $G$ with $|M| = 55$, giving us $G = MN$. Since $M$ and $N$ are abelian and $N$ is central, $G$ is abelian.
