# Group and subgroup

• Apr 5th 2010, 04:15 PM
bhitroofen01
Group and subgroup
(G;x) is a group ( e it's identity element). $\displaystyle H_1$ and $\displaystyle H_2$ are a subgroups from (G,x).
$\displaystyle H_1H_2$={$\displaystyle x_1x_2$/$\displaystyle (x_1;x_2$}$\displaystyle \in$ $\displaystyle H_1$x$\displaystyle H_2$}

1)-I must show that : $\displaystyle H_1H_2$ is a subgroup from(G;x) $\displaystyle \Longrightarrow H_1H_2=H_2H_1$
2)- We assume that $\displaystyle H_1$ and $\displaystyle H_2$ are finished and $\displaystyle H_1\cap H_2$={e} and:
$\displaystyle \phi$ : $\displaystyle H_1$x$\displaystyle H_2$ $\displaystyle \longrightarrow $$\displaystyle H_1$$\displaystyle H_2$
($\displaystyle x_1$;$\displaystyle x_2$)$\displaystyle \longrightarrow$ $\displaystyle x_1$$\displaystyle x_2 I must show that \displaystyle \phi is surjective and Card(\displaystyle H_1H_2)=Card(\displaystyle H_1)Card(\displaystyle H_2) I don't know anything!!!!!!! Can you give me some help please???? • Apr 5th 2010, 04:20 PM Drexel28 Quote: Originally Posted by bhitroofen01 Let \displaystyle H_1,H_2\leqslant G be such that \displaystyle H_1\cap H_2=\{e\}. Define \displaystyle \phi:H_1\times H_2\to H_1H_2 by \displaystyle (h_1,h_2)\overset{\phi}{\longmapsto}h_1h_2. Prove that \displaystyle \phi is surjective. Is the above the second part? • Apr 5th 2010, 04:23 PM bhitroofen01 yes • Apr 5th 2010, 04:25 PM Drexel28 Quote: Originally Posted by bhitroofen01 yes Let's see some work. • Apr 5th 2010, 04:31 PM bhitroofen01 \displaystyle H_1$$\displaystyle H_2$ is a subgroup from (G;x) $\displaystyle \Longleftrightarrow$ $\displaystyle H_1$x($\displaystyle H_2)^{-1}\in H_1H_2$
• Apr 5th 2010, 08:31 PM
FancyMouse
Did you drop any assumptions? The statement doesn't seem true to me.