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If G is a finite group, let H be a subgroup of G, and N a normal subgroup of G, and |N| and |(G:H)| are relatively prime, then show that H is a subset of N if and only if |H| divides |N|.
I'm pretty lost in this problem, would I have to use the property of module here?
Thank you.
This is kinda similar to this problem.Quote:
Originally Posted by tttcomrader
Since $\displaystyle N$ is normal subgroup form factor group $\displaystyle G/N$.
Let $\displaystyle a\in H$ and consider $\displaystyle aN \in G/N$.
Let $\displaystyle k$ be the order of $\displaystyle aN$ in $\displaystyle G/N$.
By Lagrange's theorem we know $\displaystyle k | (G:N)$.
Also $\displaystyle (aN)^{|H|} = (a^{|H|}N) = N$ thus $\displaystyle k||H|$.
Now if $\displaystyle |H|$ divides $\displaystyle |N|$ then $\displaystyle (G:N)$ divides $\displaystyle (G:H)$.
Thus, $\displaystyle k|(G:N) \implies k|(G:H)$
Thus, $\displaystyle k|H| \implies k||N|$.
But $\displaystyle (G:H)$ and $\displaystyle |N|$ are relatively prime.
This forces $\displaystyle k=1$.
Finally since the order of $\displaystyle aN$ in $\displaystyle G/N$ is one it must mean $\displaystyle a\in N$.
Therefore, $\displaystyle H\subseteq N$.
The things I don't understand here are that, if $\displaystyle |H| | |N| $, how do I know that $\displaystyle | (G:H) | | | (G:H)| $?
Also, which theorem states that $\displaystyle |G| / |N| = [G:N] $? Does it only works when N is a normal subgroup?
Thanks!
There are $\displaystyle (G:N)$ cosets and each one has $\displaystyle |N|$ elements. Furthermore, each is disjoint and together (in union) give $\displaystyle G$. Thus, $\displaystyle (G:N)|N| = |G|$.Quote:
Also, which theorem states that $\displaystyle |G| / |N| = [G:N] $? Does it only works when N is a normal subgroup?
Thanks!
Since $\displaystyle |H| | |N|$ it means $\displaystyle \frac{|N|}{|H|}$ is an integer.Quote:
Originally Posted by tttcomrader
This means $\displaystyle \frac{(G:H)}{(G:H)} = \frac{|G|/|H|}{|G|/|N|} = \frac{|N|}{|H|}$ is an integer.
