# <a>=<a^-1>

• October 21st 2009, 11:24 AM
GTK X Hunter
<a>=<a^-1>
Let $H$ be subgroup of $G$, show that
i) $a\notin H \Rightarrow H\cap Ha=\emptyset$
ii) $b\notin Ha \Rightarrow Ha\cap Hb = \emptyset$
• October 22nd 2009, 10:52 AM
proscientia
We will prove the contrapositive statements.

(i)

Suppose $x\in H\cap Ha.$ Then $x=h_1=h_2a$ for some $h_1,h_2\in H,$ $\therefore\ a=h_2^{-1}h_1\in H.$

(ii)

Suppose $x\in Ha\cap Hb.$ Then $x=h_1a=h_2b$ for some $h_1,h_2\in H,$ $\therefore\ b=h_2^{-1}h_1a\in Ha.$