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I can't solve this problem, please help me!

Every group of order 105 is isomorphic to $\mathbb{Z}_5\times H$ where $H$ is a group of order 21.

Creyesm
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  • What have you tried? And what do you know about groups? Do you know Cauchy's theorem for example? – Mathmo123 Jul 14 '14 at 00:57
  • Hello Mathmo123, I know the Sylow theorems and Cauchy theorem, I read them in Dummit & Foote book. – Creyesm Jul 14 '14 at 00:59
  • What have you tried to solve this problem? Do you see why, for example, $\mathbb Z_5$ is a subgroup? – Mathmo123 Jul 14 '14 at 01:01
  • $5$ is a prime which divides $105$, so by Cauchy Theorem exist an element $j$ of order $5$ in this group. This group is cyclic because is of prime order, so the subgroup generated by $p$ is isomorphic to $\mathbb{Z}_5$. That's what I've done but I don't know how to construct the group of order 21. – Creyesm Jul 14 '14 at 01:05

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Note that $105 = 3\cdot 5\cdot 7$. Now think about Sylow's theorems. If $n_7=1$, we're done since we have a subgroup of order $3\times 7$. If $n_7=15$, we have $15\times 6=90$ elements of order $7$. We're then left with only $15$ elements. But $n_5=1,21$. So $n_5=1$. But then we have only one $3$-Sylow, and hence we again have a subgroup of order $3\times 7$. It remains that you show that if $H$ is a subgroup of $G$ of order $21$, $H\lhd G$, and that $n_5=1$.

(Recall that if $H\lhd G$ and $K\leqslant G$, $HK\leqslant G$, and $|HK|=|H||K|/|H\cap K|$.)

Pedro
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