I Cosets: Is [gH] a Subgroup of G?

[Clear Mind]

Hi,
this week I've started to study group theory and there's a thing that i don't understand about cosets: Suppose to have a group $\textit{G}$ and a subgroup $\textit{H}$, than $\forall g \in \textit{G}$ i can build my left coset $[g \textit{H}]$. Now, here's my question: Once you have chosen an element $g_{1} \in \textit{G}$, does my left coset $[g_{1} \textit{H}]$ form a subgroup of $\textit{G}$ or it's just a subset?
Thanks in advance for the help!

 

[Stephen Tashi]

Once you have choose an element $g_{1} \in \textit{G}$, does my left coset $[g_{1} \textit{H}]$ form a subgroup of $\textit{G}$ or it's just a subset?

Always check the trivial cases. If you pick $g_1$ to be the identity element, the coset $g_1H = H$, so it is possible for a coset to be a group. In fact, what happens when you pick $g_1$ to be any particular element of $H$ ?

The interesting cases will be when $g_1$ is not an element of $H$. If you try an example, you'll see that the coset $g_1H$ is not a subgroup of G because it does not contain the identity element of $G$. Can you prove this defect always happens ?

 

[Clear Mind]

In fact, what happens when you pick $g_1$ to be any particular element of $H$ ?

Well, if $g_1 \in H$ than i suppose that $\forall g_1$ and $\forall h \in H , g_1*h \in H$ so my left coset $[g_1 H]$ it's a subgroup.

Can you prove this defect always happens ?

Mumble ... i think that if $g_1 \not\in H$ than $\forall h \in H$, $\tilde{g_1}=g_1 * h \not\in [g_1 H]$ (but $\tilde{g_1} \in G$); so if you take the element $\tilde{g}_1^{-1}$ you will have that $\tilde{g_1} * \tilde{g}_1^{-1} = \mathbb{I}$, where $\mathbb{I} \not\in [g_1 H]$. But my math skills are quite horrible, so i don't know if it's correct :(

 

[Stephen Tashi]

Well, if $g_1 \in H$ than i suppose that $\forall g_1$ and $\forall h \in H , g_1*h \in H$

That shows $[g_1H] \subset H$. You should also show that $H \subset [ g_1H]$

Mumble ... i think that if $g_1 \not\in H$ than $\forall h \in H$, $\tilde{g_1}=g_1 * h \not\in [g_1 H]$

No.
$g_1*h \in [g_1 H]$ by definition of $[g_1 H]$. The question is whether the identity of $G$ is an element of $[g_1 H]$.

Try proof by contradiction. Assume $g_1 \not\in H$ and $\mathbb{I} \in [g_1H]$. Then, for some $h \in H$, $g_1 h = \mathbb{I}$. In a group, inverses are unique and left inverses are equal to right inverses. So $h^{-1} = g_1$. Since $h \in H$ and $H$ is a subgroup, $h^{-1}$ is also an element of $H$.

 

[Clear Mind]

That shows $[g_1H] \subset H$. You should also show that $H \subset [ g_1H]$

I've tryed to prove it, but i didn't succeed. By the way i think I've found a more directly way to show that $H=[g_1 H]$ if $g_1 \in H$, but I'm stuck:
Let's ($\forall g_1 , g_2 \in H$) build two left cosets, $[g_1 H]$ and $[g_2 H]$, now, suppose that $\exists \tilde{g}$ so that $\tilde{g} \in [g_1 H]$ and $\tilde{g} \in [g_2 H]$. Than:
$$\tilde{g}=g_1 h_1=g_2 h_2 \rightarrow g_2^{-1} g_1 h_1=h_2 \rightarrow g_2^{-1} g_1=h_2 h_1^{-1}=h $$
but that mean that $\exists h \in H : g_1=g_2 h$, so $[g_1 H]=[g_2 h H]=[g_2 H]$, the two cosets are the same coset. And here is where I'm stuck, i should now show that since $\forall g_1 , g_2 \in H , [g_1 H]=[g_2 H]=H$ ...

No.
$g_1*h \in [g_1 H]$ by definition of $[g_1 H]$.

Well ... at least I've prove that I'm an idiot! My apologies.

Try proof by contradiction. Assume $g_1 \not\in H$ and $\mathbb{I} \in [g_1H]$. Then, for some $h \in H$, $g_1 h = \mathbb{I}$. In a group, inverses are unique and left inverses are equal to right inverses. So $h^{-1} = g_1$. Since $h \in H$ and $H$ is a subgroup, $h^{-1}$ is also an element of $H$.

Ok, I've understand, by assuming that $g_1 \not\in H$ and $\mathbb{I} \in [g_1H]$ you end up finding the contradiction: $g_1 \not\in H$ and $g_1=h^{-1} \in H$, where $h^{-1}$ must $\in H$ by the definition of group.
Many thanks for the help!

 

[Stephen Tashi]

so $[g_1 H]=[g_2 h H]=[g_2 H]$

If you are assuming $[hH] = H$ then you're assuming what was to be proven.

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For two arbitrary subsets of a group $G$, one may define the product of those two sets. A coset is a special case of this where one set is a singleton and the other set is a subgroup. One may ask what special properties cosets have that arbitrary subsets of group need not have. Two cosets of the same subgroup $H$ are either identical or they are disjoint. The work you are doing considers two cosets $[g_1H]$ and $[g_1H]$, so some modification of your proof might get you that result.

The collection of distinct cosets of $H$ is a collection of disjoint subsets of $G$, so it's natural to ask whether it is a partition of the set $G$.

If we have collection of subsets of $G$ , then it's interesting to ask if these sets can be regarded as elements of another group where the group operation is "take the set product". Cosets of the same subgroup don't always form a group when regarded in this manner. For special types of subgroups ("normal" subgroups) their cosets do form a group.