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

[NoName3]

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!

 

[Deveno]

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 $Hg' \subseteq Hg$?

 

[NoName3]

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 $Hg = Hg'.$

 

[Deveno]

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).