Let $G$ be an abelian finite group and let $G^{(p)}$ be its unique $p$-Sylow subgroup. Is there a way to write $G^{(p)}$ as a quotient of $G$, i.e. is there a subgroup $H\subseteq G$ with $G/H\cong G^{(p)}$?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Use the structure theorem of finite abelian groups and the fact that $C_{mn}\simeq C_m\times C_n$ whenever $\gcd(m,n)=1$. $\endgroup$– Jyrki LahtonenOct 19, 2016 at 16:31
-
$\begingroup$ Lovely, I will add that to Simon's answer. $\endgroup$– Lukas D. SauerOct 19, 2016 at 16:37
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Yes. A finite abelian group is always the direct product of its Sylow subgroups. I.e. $\begin{align}G\cong G^{(p_1)}\times G^{(p_2)}\times...\times G^{(p_n)}\end{align}$ where $p_i$ are the distinct primes in the factorization of $|G|$. This follows from the structure theorem for finite abelian groups. So you can simply choose $H$ as the product of all other Sylow subgroups.