Show that left cosets partition the group

Solution 1:

Here is the proof for (b)

suppose that there is an element $c \in gH \cap g^\prime H$ so that means that $c \in gH$ and $c \in g^\prime H$ then this means that there are elements $a,b \in H$ such that $c = ga$ and $c = g^\prime b$

now this implies that $$c=ga=g^\prime b$$ which also implies that $$g = g^\prime b a^{-1}$$ because we can right multiply by $a^{-1}$ on each side.

However, observe that $ba^{-1} \in H$ (Closure of groups)

and so $$g \in g^\prime H$$

and so for any $a \in H$, $ga \in g ^\prime H$

which implies that $$gH \subseteq g^\prime H$$ By a symmetrical argument ,$$g^\prime H \subseteq gH$$

and hence $$gH = g^\prime H$$

Your argument for part (a) Looks decent :)

Solution 2:

Part (1) looks perfectly fine.
For part (2), assume that $g_1H\cap g_2 H\ne(e).$

Then it follows that, for some $h_1,h_2\in H, g_1h_1=g_2h_2$. Then $g_1h_1h_2^{-1}=g_2$. But since $h_1h_2^{-1}\in H$, it follows that $g_2\in g_1H$. So $g_2=g_1h_3$ for some $h_3\in H$. Thus $h_3=g_1^{-1}g_2$. And so, using the closure of $H$, all elements of the form $g_1^{-1}g_2h$ are contained in $H$, and so all elements of the form $g_1g_1^{-1}g_2h=g_2h$ are contained in $g_1H$. And so you can conclude that $g_2H\subseteq g_1H$. Using similar logic gives you that $g_1H\subset g_2H$, and hence, $g_1H=g_2H$.