TITLE: Show that left cosets partition the group
QUESTION [11 upvotes]: I know how to prove that it happens, by proving that the left coset definition actually is an equivalence relation. Then, it's proved that it partitions the set, since equivalence relations do it.
However, this exercise asks me to prove it in a diferente way. First it asks me to show that:
a) The union of the left cosets is equal to $G$
b) If $gH \cap g'H \neq \emptyset$ then $gH = g'H$
for $a)$ I'm  thinking about the following:
$gH = \{gh, h\in H\}$
So if $G = \{g_1, g_2, \cdots g_n\}$
We would have the following left cosets:
$$g_1H = \{g_1h, h\in H\}$$
$$g_2H = \{g_2h, h\in H\}$$
$$\cdots$$
$$g_nH = \{g_nh, h\in H\}$$
The union of all these sets will include all the $g's$, since for each set
$$g_k = \{g_kh, h\in H\}$$
we have
$$ge \in g_k = \{g_kh, h\in H\}$$
where $e$ is the identity.
Then if we make the union of all these sets we'll have at least all the elements of $g$. The other elements are merely $gh$ for some $h$. But since $gh\in G$ they would be repeated elements in the union, doesn't matter. So, the union of all left cosets of $H$ in $G$ is $G$.
Is my reasoning correct?
Also, what could I do to prove $b)$?

REPLY [9 votes]: 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 :)