# Question involving Sylow's Thm for $|G|=245$

Let $G$ be a group of order $245=7^2\cdot 5$.

1) Use Sylow's Thm to show has normal subgroup $H$ of order $49$.

So by definition of a $p$-Sylow subgroup, where $7$ does not divide $5$, then for $H$ is a subgroup of $G$. By Sylow's Thm, $n_7\mid 5$ or $n_7=1\bmod(5)$. So there is only one $7$-sylow subgroup in $G$. Therefore $H$ is normal?? Also how would I show that $H$ has order $49$? Lastly, why is $H$ abelian?

2) Show it has a normal subgroup of order $5$.

Im assuming this will be the same as above.

3) Show $G$ is abelian. What are the only possible isomorphism classes for any group $G$ of order $245$.

I don't really see how $G$ could be abelian from previous work, but if the possible classes for $G$ are either $Z_{245}, Z_{49}+Z_5, Z_{35}+Z_7, Z_7+Z_7+Z_5$ where $+$ means direct product. Each of these are abelian since they are cyclic. Would that be enough?

• Do you mean "I don't really see how G could be nonabelian" ? – Peter Nov 29 '14 at 18:46
• sorry, I meant as in from previous work. I could see it since it makes some sense, but of course thats never really a viable answer – Jack Armstrong Nov 29 '14 at 18:48
• I'm having trouble following you. The groups you mention are not "abelian since they are cyclic"; they are obviously abelian because they are the direct products of cyclic groups (which are themselves abelian). – Hew Wolff Nov 29 '14 at 19:09
• I think Sylow's Theorem Application. Prove G is Abelian has most of what you want. – Hew Wolff Nov 29 '14 at 19:13
• everything but normal subgroup of order 5 – Jack Armstrong Nov 29 '14 at 19:58

Let $H \in Syl_7(G)$. And $K \in Syl_5(G)$. You have shown correctly that in fact $H$ and $K$ are the only Sylow $7$-subgroup, and Sylow $5$-subgroup, respectively. So both subgroups $H$, and $K$ are normal. Also, both groups are abealian. $K$ is cyclic of order $5$. And for $H$ use the fact that in general, a group of order $p^2$, with $p$ a prime, must be abelian.
Moreover, $H \cap K=1$ and $G=HK$ (this last fact can be seen by counting: $|HK|=\frac{|H| \cdot |K|}{|H \cap K|}$). It now follows that $G=H \times K$. Since both $H$ and $K$ are abelian, $G$ must be abelian. And going one step further either $G \cong C_{245}$ or $\cong C_5 \times C_7 \times C_7$ (the last one is isomorphic to $C_{35} \times C_7$).