Prove the set of all left (right) cosets of $H$ partitions $G$.

Given a group $$(G,\circ)$$ and $$H \le (G,\circ)$$,the left coset of $$H$$ is the set of all elements of $$H$$ multiplied by a fixed element in $$G$$,formally given $$g \in G$$,then the left cosets of $$H$$ is denoted by $$gH$$ and is defined as:$$gH:=\left\{gh:h \in H \right\}$$

Similarily the right coset of $$H$$ is defined.

Theorem: Prove the set of all left (right) cosets of $$H$$ partitions $$G$$.

A partition of a given set is a family of nonempty subsets that are pairwise disjoint and their union is the whole set.

It's nedeed to show that for every two distinct left cosets $$g_1H$$ and $$g_2H$$ they don't have any element in common.For the sake of contradiction assume $$g_1H \ne g_2H$$ but $$g_1H \cap g_2H \ne \emptyset$$,equivalently there exist $$x$$ which is in both of them.

By the definition:

$$x=g_1h_1\;\;\text{for some}\;\; h_1 \in H \;\;\text{and}\;\; x=g_2h_2 \;\;\text{for some}\;\; h_2 \in H$$

Hence $$g_1h_1=g_2h_2$$,$$H$$ is a group and this ensures the existese of $$h_{1}^{-1}$$,multiplying both sides of the equation by the inverse gives $$g_1=g_2h_2 \circ h_{1}^{-1}$$,closure of $$H$$ implies $$h_2 \circ h_{1}^{-1} \in H$$,this means $$g_1 \in g_2H$$ and implies $$g_1H=g_2H$$,contradicts the assumption.

On the other hand, since $$H$$ is a subgroup, hence it's a group and does have an identity element which can be shown to be the same as the identity element of $$G$$ denoted by $$e$$, from here taking $$h=e$$ follows for any fixed $$g \in G$$: $$gH \ne \emptyset$$.

It's left to show that the union of all left (right) cosets are $$G$$, this is where do I have a problem with.

It's clear that every element in $$\bigcup_{g \in G} gH$$ is an element of $$G$$, on the other hand for every $$g \in G$$ :$$g=ge \in gH$$, which means every element in $$G$$ is in the corresponding left coset and hence is the union of the corresponding cost with the other left cosets.

I think from the definition of set equality it's true to conclude that $$G=\bigcup_{g \in G} gH$$

And hence left(right) cosets of $$H$$ partition the set G.

How much of my work is true?

• if $h\in H$, then $h\in hH$. Oct 7 '20 at 13:21

(Just in case you'd be interested to a different approach.)

The relation $$a\sim b\stackrel{(def.)}{\iff}ab^{-1}\in H$$ is an equivalence relation in $$G$$. The equivalence class of $$a$$ is precisely:

\begin{alignat}{1} [a]_\sim &= \{b\in G\mid b\sim a\} \\ &= \{b\in G\mid ba^{-1}\in H\} \\ &= \{b\in G\mid b\in Ha\} \\ &= Ha \end{alignat}

As equivalence classes, the cosets of a subgroup always partition the group.

Specifically for you question, final part, note that for every $$\tilde g\in G$$, we get $$\tilde g\in H\tilde g\subseteq \bigcup_{g\in G}Hg$$ and hence $$G\subseteq \bigcup_{g\in G}Hg$$, which is basically what you found out by yourself. So, that's okay. Also for the first part, I see no flaws; rather, the only I would mention is that the implication $$g_1\in g_2H\Rightarrow g_1H=g_2H$$ uses the surjectivity of the map $$H\to H, h\mapsto \tilde hh$$, for any given $$\tilde h\in H$$:

\begin{alignat}{1} g_1\in g_2H &\Rightarrow \exists\tilde h\in H\mid g_1=g_2\tilde h \\ &\Rightarrow g_1H =\{g_1h,h\in H\} \\ &\Rightarrow g_1H =\{g_2\tilde hh,h\in H\} \\ &\Rightarrow g_1H =\{g_2h',h'\in H\} \\ &\Rightarrow g_1H =g_2H \\ \end{alignat}

• @ user750041,So nice and thanks for your answer, It would be appreciated if you verify my solution too. Oct 7 '20 at 14:23
• It seems okay to me, @45465. I've just added a comment to my answer.
– user810157
Oct 7 '20 at 15:17