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If $|G|=65$ then $G$ contains an element of order $5$.

Proof: If $G$ is cyclic then there is some $x \in G$ where $|x| = 65$ so $|x^{13}|=5$.

Assume $G$ is not cyclic and that no elements in $G$ have order $5$. By LaGrange's theorem all elements of $G$ have their order dividing $G$ so every non-identity member of $G$ has order $13.$

(From here I have seen the solution and the claim at this point is that if $x \neq y$ then $\left< x \right> \cap \left< y \right> = \{ e \}.$ The trouble I am having is why this is true. The hint is that the order of $\left< x \right> \cap \left< y \right>$ divides both the order of $\left< x \right>$ and the order of $\left< y \right>$. If the intersection of the cyclic groups had order $13$, this would somehow imply that $x=y$? I have seen in the symmetry group $S_3$ that different elements can give the same cyclic group. I am sort of unsure why this is true or maybe I am missing something else.)

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Choose $y \notin \langle x \rangle$, which we can do since $|x| = 13$. Then $\langle y \rangle \cap \langle x \rangle = 1$ because the only alternative is that $|\langle y \rangle \cap \langle x \rangle| = 13$, implying $\langle x \rangle = \langle y \rangle$. So every non-identity element is in exactly one cyclic group of order 13, which leads to a quick contradiction with $|G| = 65$.

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Of course it's not true. If $x$ has order 13, then each of its powers different from $e$ generates $\langle x \rangle$.

What must have been claimed is that two distinct subgroups of order 13 have trivial intersection. (The argument you sketch, which involves Lagrange, points in that direction.)

This allows you to conclude, by counting the elements of the group in dozens, plus the identity.

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Okay yes that sounds right. Okay thank you. –  Starlight Mar 7 '13 at 7:40
    
@Wishingwell, you're welcome. –  Andreas Caranti Mar 7 '13 at 7:41
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You've started out quite nicely. Trying to pick up from where you're stuck, and given the clarification of what must have been intended, or left out, by the assertion you're puzzled by:

Recall that if the only subgroups $H_i$ of $G$ are of order $13$, since $13$ is prime, each non-identity element in $H_1, H_2, ...$ must generate a subgroup of order $13$. That is, for all $x \in G, x \neq e$, then $|\langle x\rangle| = 13$ and for any given $x$, every non-identity element in $\langle x \rangle$ generates $\langle x \rangle$ too. So if $y \in G, y\notin \langle x\rangle,$ then $\langle x \rangle \cap \langle y \rangle = \{e\}$.

As Andreas suggests, you can then conclude that subgroups each contain 12 elements distinct from elements in other groups, with one element, $e$, common to them all. So say there are $n$ distinct subgroups of order $13$, then $|G| = 65 \neq 12n + 1$, for any $n$. Hence if G is not cyclic, then it cannot be the case that every subgroup of $G$ has order $13$.

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I like this thank you –  Starlight Mar 7 '13 at 8:06
    
You're very welcome ;-) –  amWhy Mar 7 '13 at 8:21
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