<a>=<a^-1>

**GTK X Hunter** · October 21st 2009, 10:24 AM

Let $H$ be subgroup of $G$ , show that
i) $a\notin H \Rightarrow H\cap Ha=\emptyset$
ii) $b\notin Ha \Rightarrow Ha\cap Hb = \emptyset$

**proscientia** · October 22nd 2009, 09:52 AM

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 $h_1,h_2\in H,$ $\therefore\ b=h_2^{-1}h_1a\in Ha.$

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