Let $G$ be a group of order $35$. Prove that $G \cong C_5 \times C_7$.
$G$ has subgroups of orders $5$ and $7$ by Lagrange's theorem? If so, call them $A$ and $B$. I know their intersection is $\{1_G\}$. But how do I include the cyclic groups?
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The only group of any prime order $p$ is the cyclic group $C_p$. – Travis Willse Feb 11 '16 at 22:41
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1In standard group theory terminology $G * H$ means the free product of $G$ and $H$. You should write $G \times H$ for the direct product. (I have fixed this for you above.) – Rob Arthan Feb 11 '16 at 22:46
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Thanks. So the subgroups A and B are cyclic and $A \cong C_5$ and $B \cong C_7$ ? – George Feb 11 '16 at 22:48
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Note that $C_5 \times C_7 \cong C_{35}$, so you have to prove that the only group with order 35 is the cyclic group. Now, this Wikipedia page may help. – Crostul Feb 11 '16 at 22:51
2 Answers
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Apply Sylow's theorem: let $v_p$ be the number of sylow $p$-subgroups.
Then $v_5$ divides $7$ and $v_5\equiv 1\bmod 5$. So $v_5=1$ (this implies the unique subgroup of order $5$ is normal).
We also have $v_7$ divides $5$ and $v_7\equiv 1\bmod7$. So $v_7=1$ (this implies the unique subgroup of order $7$ is normal).
So we have a subgroup $H$ of order $5$ that is normal and a subgroup $K$ of order $7$ that is normal (we use Cauchy's theorem to assert this).
We now use $|HK|=\frac{|H||K|}{|H\cap K|}$, since $H$ and $K$ have coprime order by Lagrange's theorem we have $|H\cap K|=1$.
So we conclude $|HK|=35$, so $HK=G$.
We now invoke the following theorem: If $H$ and $K$ are normal subgroups of $G$ such that $H\cap K=1$ and $G=HK$ we have $G\cong H\times K$.
It is clear $H\cong C_5$ and $K\cong C_7$
Asinomás
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if a subgroup is the only subgroup of its order then it is normal (because $gHg^{-1}$) is a subgroup with the same order as $H$) – Asinomás Feb 12 '16 at 00:12
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Why do we need Cauchy's theorem? Isn't it true from the implications above – George Feb 12 '16 at 19:06
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Proof sketch:
By Sylow's theorems, one checks the (cyclic) subgroup $B$ of order $7$ is normal in $G$. If $A$ is a subgroup of order $5$, necessarily cyclic, one deduces from $A\cap B=\{e\}$ and $\lvert A\times B\rvert=\lvert G\rvert$ that $G=B\times A$ is isomorphic to a semi-direct product $$\mathbf Z/7\mathbf Z\rtimes_\theta\mathbf Z/5\mathbf Z,$$ where $\theta: \mathbf Z/5\mathbf Z\longrightarrow \operatorname{Aut}(\mathbf Z/7\mathbf Z)$ is a group homomorphism.
Now $ \operatorname{Aut}(\mathbf Z/7\mathbf Z)$ has order $6$, so $\theta(1)$ has order a divisor of $6$, and also a divisor of $5$, hence $\theta(1)=\operatorname{id}_{\mathbf Z/7\mathbf Z}$, which proves the semi-direct product is the ordinary direct product, so the group $G$ is commutative, isomorphic to $\mathbf Z/35\mathbf Z$.
Bernard
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