Proving Equivalence of $g' \in Hg$, $g'g^{-1} \in H$, and $Hg = Hg'$

  • MHB
  • Thread starter NoName3
  • Start date
  • Tags
    Equivalence
In summary, we are given a subgroup $H$ of a group $G$, and elements $g'$ and $g$ of $G$. We are asked to prove that the following statements are equivalent: $(a)$ $g' \in Hg$, $(b)$ $g'g^{-1} \in H$, and $(c)$ $Hg = Hg'$. Using the fact that $Hg$ and $Hg'$ are cosets of $H$, we can show that $(a) \implies (b)$ and $(b) \implies (c)$. To show $(c) \implies (a)$, we use the assumption that $Hg = Hg'$ and the
  • #1
NoName3
25
0
Let $H$ be be a subgroup of a group $G$. Let $g'$ and $g$ elements of $G$. Prove that the following are equivalent: $(a)$ $g' \in Hg$, $(b)$ $g'g^{-1} \in H$, and $(c)$ $Hg = Hg'$.

$g' \in Hg$ means $g' = hg$ for some $g \in G$ and $h \in H$. And $g' = hg \implies g'g^{-1} = hgg^{-1} = h$. But $h \in H$ so $g'g^{-1} \in H$.

So $(a) \implies (b)$. I can't progress. For a moment I thought I had it, then I lost it!
 
Last edited:
Physics news on Phys.org
  • #2
Suppose $gg'^{-1} \in H$, say:

$gg'^{-1} = h$.

Then $g = hg'$. Does it not follow that for any other $h_1 \in H$, that:

$h_1g = h_1(hg')$?

Use this to show $Hg \subseteq Hg'$.

Can you show likewise that $Hg' \subseteq Hg$?
 
  • #3
Deveno said:
Suppose $gg'^{-1} \in H$, say:

$gg'^{-1} = h$.

Then $g = hg'$. Does it not follow that for any other $h_1 \in H$, that:

$h_1g = h_1(hg')$?

Use this to show $Hg \subseteq Hg'$.

Can you show likewise that $Hg' \subseteq Hg$?
Thank you.

So let $x = h_1g$. Then $x = h_1hg' \in H$ because since $h_1, h \in H$ we have $h_1 h \in H$. Therefore $Hg \subseteq Hg'$. For the converse, suppose $h_2 \in H$ and let $y = h_2 g' = h_2h^{-1}g \in H$ because since $h_2 , h^{-1} \in H$ we have $h_2 h^{-1} \in H$. Therefore $Hg' \subseteq Hg$. Hence $Hg = Hg'.$
 
Last edited:
  • #4
NoName said:
Thank you.

So let $x = h_1g$. Then $x = h_1hg' \in H$ because since $h_1, h \in H$ we have $h_1 h \in H$. Therefore $Hg \subseteq Hg'$. For the converse, suppose $h_2 \in H$ and let $y = h_2 g' = h_2h^{-1}g \in H$ because since $h_2 , h^{-1} \in H$ we have $h_2 h^{-1} \in H$. Therefore $Hg' \subseteq Hg$. Hence $Hg = Hg'.$

Another way to do this is using the fact that either:

a) $Hg = Hg'$ -or-
b) $Hg \cap Hg' = \emptyset$

in which case it suffices to show that $Hg \cap Hg' \neq \emptyset$ to show the two cosets are the same.

Why are a) and b) true?

Because:

$Hg = Hg' \iff gg'^{-1} \in H$, and:

$g \sim g' \iff gg'^{-1} \in H$ is an EQUIVALENCE RELATION on $G$ (called "congruence modulo $H$").

You have seen this before, although you probably did not recognize it at the time.

The Euclidean plane:

$\{(x,y):x,y \in \Bbb R\}$

is a group, under the operation:

$(x,)\ast(x',y') = (x+x',y+y')$ (the normal vector addition).

A line through the origin, of slope $m$, is a SUBGROUP:

$L_m = \{(x,y) \in \Bbb R^2: y = mx\}$ (this is all points of the form $(x_0,mx_0)$).

The slope-intercept form of a line:

$y = mx + b$

Is a "congruence class" of $L_m$, since if we take two such points, and subtract (this is $pp'^{-1}$ for our 2-vectors and the operation of vector addition): we have:$(x_2,mx_2 + b) - (x_1,mx_1 + b) = (x_2-x_1,m(x_2-x_1)) \in L_m$.

So we think of the line $y = mx + b$ as "$L_m$ translated (up) by $b$", in much the same way:

$Hg$ is $H$ "translated" (multiplicatively) by $g$.

In other words, lines not through the origin are cosets of a parallel line through the origin (thinking about cosets this way makes Lagrange's Theorem make more "sense", because we see cosets are in some sense, "parallel", like equal slices of a rectangular cake), to see if two group elements are "in the same slice" we do the group analogy of subtraction, which is, evaluate $gg'^{-1}$, and see if it lies in the "home slice" (the one that contains the identity).
 

FAQ: Proving Equivalence of $g' \in Hg$, $g'g^{-1} \in H$, and $Hg = Hg'$

What does it mean to prove equivalence in this context?

Proving equivalence means demonstrating that two statements or concepts have the same meaning or are equivalent to each other.

Why is proving equivalence important in this scenario?

In this scenario, proving equivalence is important because it helps establish the relationship between different elements in a group, which can then be used to make further deductions and prove mathematical theorems.

How can we prove that $g' \in Hg$ is equivalent to $g'g^{-1} \in H$?

We can prove this equivalence by showing that if $g' \in Hg$, then $g'g^{-1} \in H$, and vice versa. This can be done using definitions and properties of groups, such as the closure property and the fact that every element in a group has an inverse.

Is proving $Hg = Hg'$ the same as proving $g' \in Hg$ and $g'g^{-1} \in H$?

Yes, proving $Hg = Hg'$ is equivalent to proving both $g' \in Hg$ and $g'g^{-1} \in H$. This is because if $Hg = Hg'$, then $g' \in Hg$ and $g'g^{-1} \in H$ must both be true, and vice versa.

Can we use the concept of equivalence to prove other properties of groups?

Yes, the concept of equivalence can be used to prove many other properties of groups and other mathematical structures. For example, we can use equivalence to prove that two groups are isomorphic, meaning they have the same structure, or to show that two operations in a group are equivalent.

Similar threads

Replies
1
Views
1K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
1
Views
7K
Replies
6
Views
1K
Replies
5
Views
2K
Replies
3
Views
1K
Back
Top