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

• I
• Clear Mind
In summary, the conversation discusses cosets in group theory and whether they form subgroups or just subsets of a group. The speaker provides examples and attempts to prove the properties of cosets. The expert suggests using proof by contradiction and mentions the special properties of cosets and their potential to form a partition or a group under certain conditions.

#### 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!

Last edited:
Clear Mind said:
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 ?

Stephen Tashi said:
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.

Stephen Tashi said:
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 :(

Clear Mind said:
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##.

Stephen Tashi said:
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## ...

Stephen Tashi said:
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.

Stephen Tashi said:
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!

Clear Mind said:
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.

-------

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.

## 1. What is the definition of a coset?

A coset is a subset of a group G that is obtained by multiplying each element of a subgroup H by a fixed element g in G.

## 2. How do you determine if [gH] is a subgroup of G?

To determine if [gH] is a subgroup of G, you must first check if gH is a subset of G, and then verify that gH is a group under the same operation as G.

## 3. What is the significance of cosets in group theory?

Cosets are important in group theory because they allow us to partition a group into smaller subgroups, which can make it easier to analyze and understand the structure of the group.

## 4. Can a coset be a subgroup of G if H is not a subgroup of G?

No, a coset can only be a subgroup of G if H is also a subgroup of G. If H is not a subgroup of G, then the coset [gH] will not satisfy the properties of a subgroup.

## 5. How do you prove that [gH] is a subgroup of G?

To prove that [gH] is a subgroup of G, you must show that it is closed under the operation in G, that it contains the identity element of G, and that it contains the inverse of each element in [gH]. This can be done by using the definition of a subgroup and properties of cosets.