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Prove that every group of order $35$ is cyclic.

Now, the subgroups of this are ones whose orders divide the order of this group(by lagrange), these are of prime orders $7$ and $5$.

and I guess $\Bbb Z_7\times \Bbb Z_5$ is of order $35$ and since these are both cyclic, so is $\Bbb Z_{35}$.

But that doesn't prove anything about every subgroup of order $35$.

How do I do this.

I cut trees
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  • Have you learned the Sylow theorems? – Marcus M Jun 21 '15 at 12:51
  • @MarcusM I haven't, but I can. – I cut trees Jun 21 '15 at 12:51
  • Okay, it's fairly straightforward with the Sylow theorems, but there might be a way to do the problem without it. There are a lot of theorems to help you out in this problem; another (slightly more obscure) theorem that would apply here is that for every group of order $n$, if $n$ and $\varphi(n)$ are coprime, then the group is cyclic. Here we have $n = 35$, $\varphi(n) = (7 - 1)(5 - 1) = 24$. Since these two numbers are coprime, the group is cyclic. – Marcus M Jun 21 '15 at 12:56

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The number of 7 Sylow subgroups divides 5 and is congruent to 1 mod $7$, hence there's a unique one, and hence normal. Similarly, the number of $5$ Sylow subgroups divides 7 and is congruent to $1$ mod $5$, hence there's a unique one. It follows that $G \cong \mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}\cong \mathbb{Z}/35\mathbb{Z}$ by Chinese remainder theorem.

Glorfindel
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mich95
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