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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?

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    Dear Alex, Suppose that $G$ is abelian; then what would its structure be? Now, try to prove that $G$ necessarily *has* this structure. What kind of subgroup of $G$ do you have to exhibit? What methods do you have available for exhibiting such a subgroup? Regards,2012-01-03
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    Sylow $11$-subgroups?2012-01-03
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    Dear Alex, Just to clarify, my hint was an oblique form of Dylan's answer. An abelian group of order $55\cdot 169$ is the direct product of a cyclic group of order $55$ and a group of order $169$. You are given the order $55$ part already (it is given that $G$ contains a cyclic subgroup of order $55$), and so you are left with the order $169$ part to deal with. It is natural to use Sylow theory to do this; more details are in Dylan Moreland's answer below. Regards,2012-01-03
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    I should rewrite the answer to better reflect this. By the way: all abelian groups of order $55$ are cyclic, I think.2012-01-03

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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.

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    Dear Dylan, I wouldn't call this artless; it is the natural way to do it, as far as I can tell (and what I was alluding to, perhaps too obliquely, in my comment above). Regards,2012-01-03
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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.

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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.

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    Interesting use of commutator subgroups.2012-05-23
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    @awllower: Thanks. In general it is true that $N$ normal and $G' \cap N = \{1\}$ implies $N$ central. This sometimes (not always) gives alternate proofs for showing that a normal subgroup is central.2012-06-02
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    Thanks again, for this extra information.2012-06-02