# another group theory question

• Nov 14th 2010, 05:03 AM
situation
another group theory question
Let G be a group and H be a subgroup of G. For any $\displaystyle g_1 , g_2 \in G$ prove that:
$\displaystyle g_2 \in g_1H \Rightarrow g_1H = g_2H$
Use this to prove that any two left cosets of H are equal or disjoint.

help would be much appreciated! thanks :)
• Nov 14th 2010, 07:48 AM
Drexel28
Quote:

Originally Posted by situation
Let G be a group and H be a subgroup of G. For any $\displaystyle g_1 , g_2 \in G$ prove that:
$\displaystyle g_2 \in g_1H \Rightarrow g_1H = g_2H$
Use this to prove that any two left cosets of H are equal or disjoint.

help would be much appreciated! thanks :)

I'm sure you're aware that $\displaystyle (ab)H=a(bH)$ and $\displaystyle hH=H$ for all $\displaystyle h\in H$. So, note that $\displaystyle g_2\in g_1 h\implies g_2=g_1h$ for some $\displaystyle h\in H$ and so $\displaystyle g_1h^{-1}=g_2$. Thus, $\displaystyle g_2H=(g_1h^{-1})H=g_1(h^{-1}H)=g_1H$
• Nov 15th 2010, 03:36 AM
situation
Quote:

Originally Posted by Drexel28
I'm sure you're aware that $\displaystyle (ab)H=a(bH)$ and $\displaystyle hH=H$ for all $\displaystyle h\in H$. So, note that $\displaystyle g_2\in g_1 h\implies g_2=g_1h$ for some $\displaystyle h\in H$ and so $\displaystyle g_1h^{-1}=g_2$. Thus, $\displaystyle g_2H=(g_1h^{-1})H=g_1(h^{-1}H)=g_1H$

thanks man, just a quick one about the result $\displaystyle hH=H$ how is this the case? if you wouldnt mind explaining that further it would be much appreciated! :)
• Nov 15th 2010, 03:49 AM
Drexel28
Quote:

Originally Posted by situation
thanks man, just a quick one about the result $\displaystyle hH=H$ how is this the case? if you wouldnt mind explaining that further it would be much appreciated! :)

No problem! By definition $\displaystyle hH=\left\{hh':h'\in H\right\}$ clearly then since each element is the multiplication of elements in $\displaystyle H$ we have that $\displaystyle hH\subseteq H$. But, if $\displaystyle h'\in H$ then $\displaystyle h^{-1}h'\in H$ and so $\displaystyle h(h^{-1}h')=h'\in hH$ and so $\displaystyle H\subseteq hH$. Therefore, $\displaystyle H=hH$ (this is a result usually proven along with Cayley's theorem, that the map $\displaystyle \sigma_h:H\to H:h'\mapsto hh'$ is a permutation)