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I am proving that a group of order $\textrm {G}$ being 65 would imply that $\textrm {G}$ is cyclic, using Lagrange's Theorem and $N \textrm{by} C$ Theorem.

Clearly, one can show that not all non-identity elements in $\textrm{G}$ would have order 13. But I have difficulty in proving that not all non-identity elements in $\textrm{G}$ would have order 5 either.

Assuming that all non-identity elements in $\textrm{G}$ have order 5, the following statement can be concluded:

1) Every non-trivial, proper subgroup of $\textrm{G}$ is a cyclic group of order 5.

2) None of those groups will be normal subgroups of $\textrm{G}$; otherwise it will imply that $\textrm{G}$ will have a subgroup of order 25 which is not possible since $|\textrm{G}| = 65$.

3) Index of every non-trivial, proper subgroup of $\textrm{G}$ will be 13.

My hunch is that point (3) would possibly show that $\textrm{G}$ doesn't have every non-identity element of order 5. But I have difficulty in showing that given any non-trivial, proper subgroup, I can construct more than 13 coset of that subgroup. Or is their any other fact that I'm missing.

Shaun
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Minto P
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    It seems to me that you are missing a very important, basic theorem: https://en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory) This shows that there must be elements with order $5$ and with order $13$, as well. – A. Pongrácz Dec 31 '18 at 09:34
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    Since $gcd(65,\phi(65))=1$, every group of order $65$ is cyclic, see this question. – Dietrich Burde Dec 31 '18 at 09:35
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    Obviously the question is easy if you know Sylow's Theorem or even Cauchy's Theorem, but presumably the whole point of the question is to avoid using either of those. One approach would be to consider the conjugation action of $G$ on the $16$ subgroups of order $5$, and use the fact that all orbit lengths must divide $65$. The only possibility if $1+5+5+5$, so there is a normal subgroup or order $5$. – Derek Holt Dec 31 '18 at 09:38

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General Result:

Let $G$ be a group and $\vert G\vert = pq$, where $p$ and $q$ are distinct primes with $p < q$ and $p \nmid q-1$. Then $G$ is cyclic.

Proof. Let $n_p$ be the number of Sylow $p$-subgroups of $G$. Then, by Sylow's Theorem, $n_p \mid q$ and $n_p = 1\pmod p$. This implies that either $n_p = 1$ or $n_p = q$. Firstly, suppose that $n_p = q$. Then $q = 1\pmod p$, and so $p \mid q-1$ - a contradiction!

Thus $n_p = 1$, and so if $P$ is a Sylow $p$-subgroup of $G$, then $P$ is unique. Thus $P \trianglelefteq G$, and by the same token $Q \trianglelefteq G$, where $Q$ is the unique Sylow $q$-subgroup of $G$. Because $\vert P\vert = p$ and $\vert Q\vert = q$, where $p$ and $q$ are primes, we can write $P = \langle x \rangle$ and $Q = \langle y \rangle$ for some $x,y\in G$ with $o(x) = p$ and $o(y)=q$. Also, since $p$ and $q$ are distinct, $P \cap Q = \lbrace 1\rbrace$ by Lagrange's Theorem. Now the commutator $[x,y] = x^{-1}y^{-1}xy \in P\cap Q = \lbrace 1 \rbrace$ (because $P$ and $Q$ are normal in $G$). Thus $x^{-1}y^{-1}xy = 1$, i.e. $xy = yx$. Then it follows easily that $o(xy) = pq$, and so $G = \langle xy \rangle$ is cyclic.

In your example, $\vert G \vert = 65 = 5\cdot 13$, so choose $p = 5$ and $q = 13$. Clearly $5 < 13$ and $5 \nmid 12$ so the hypothesis of the above result are satisfied. Then $G$ is cyclic.

Hope this helps.

For further information on Sylow's Theorems: https://en.wikipedia.org/wiki/Sylow_theorems

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