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?

  • $\begingroup$ if $h\in H$, then $h\in hH$. $\endgroup$
    – alphaomega
    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}

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

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