# show that...

• Oct 21st 2009, 10:33 AM
GTK X Hunter
show that...
Let H be subgroup of abelian group G, then show that:
i) $\displaystyle (aH)^-1=a^-1H$
ii) $\displaystyle (Ha)^2=Ha^2$
• Oct 21st 2009, 03:42 PM
aman_cc
Quote:

Originally Posted by GTK X Hunter
Let H be subgroup of abelian group G, then show that:
i) $\displaystyle (aH)^-1=a^-1H$
ii) $\displaystyle (Ha)^2=Ha^2$

Can you define $\displaystyle (aH)^-1$ and $\displaystyle (Ha)^2$ for me plz?
• Oct 21st 2009, 08:21 PM
tonio
Quote:

Originally Posted by GTK X Hunter
Let H be subgroup of abelian group G, then show that:
i) $\displaystyle (aH)^-1=a^-1H$
ii) $\displaystyle (Ha)^2=Ha^2$

Do you mean in the quotient group $\displaystyle G/H$?
This follows at once from the definitions of product in quotient groups. Read and understand well these definitions.

Tonio
• Mar 27th 2010, 12:55 AM
GTK X Hunter
take any $\displaystyle h\in H$
i) $\displaystyle (ah)^{-1}=h^{-1}a^{-1}=a^{-1}h^{-1}=a^{-1}h_o$
ii) $\displaystyle haha=hhaa=h_1a^2$
am i right? ^_^
• Mar 27th 2010, 12:41 PM
Drexel28
Quote:

Originally Posted by GTK X Hunter
take any $\displaystyle h\in H$
i) $\displaystyle (ah)^{-1}=h^{-1}a^{-1}=a^{-1}h^{-1}=a^{-1}h_o$
ii) $\displaystyle haha=hhaa=h_1a^2$
am i right? ^_^

It's simpler then that.

$\displaystyle \left(aH\right)\left(a^{-1}H\right)=(aa^{-1})H=H=e\implies a^{1}H=\left(aH\right)^{-1}$. Similarly for the second.