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I understand that the only difference between left and right group actions is the order in which gh acts on x. Since cosets Hg and gH involve the action of one element of the subgroup H on one element of G, this order difference doesn't come into play. So how do the cosets end up different in the non-normal case? I apologize if I'm wildly confused; I'm still learning!

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  • $\begingroup$ It's not clear what $x$ is in your question or even what action you're describing. But, the best way to see that left and right cosets of a subgroup can be different is by looking at an explicit example, e.g., $G$ the symmetric group $S_3$, and $H$ the subgroup generated by a transposition. Are you familiar with symmetric groups and cycle notation? $\endgroup$ Commented yesterday
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    $\begingroup$ I’m not sure I understand your question. $Hg$ and $gH$ are sets, and they do not in general have the same elements. Why do you expect them to? $\endgroup$ Commented yesterday
  • $\begingroup$ @Malady If h is an element of H, shouldn't hg (the left action) be the same as gh (the right action) since there is only one element of H acting on G? $\endgroup$ Commented yesterday
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    $\begingroup$ If the group is non-commutative, then $gh$ may not be the same as $hg$ $\endgroup$ Commented yesterday
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    $\begingroup$ If you want to act on $x$ by $f$ and then by $g$, the difference between $gfx$ and $xfg$ is only a matter of notation. But once you have decided which notation to use, then $fg$ and $gf$ are not the same, as one of them is $f$ followed by $g$, and the other is $g$ followed by $f$. $\endgroup$ Commented yesterday

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You have already been given an example, but that doesn't answer why.

The answer to, "Why?" is, "Why not!"

This isn't a snide remark. When we're dealing with noncommutative groups, we should never expect left-something to be the same as right-something unless there is some reason for it. Thus we should not generally expect the left-cosets of a subgroup to be the same as the right-cosets. Simply because there is no reason for things to work out that way.

This perspective takes a mind shift from what we're used to when everything is commutative. In most of our experience, commutativity is standard. It is expected. And it is easy to dismiss the noncommutativity of some things, like matrices and permutations, as a weird footnote. Sure, it comes up. But we don't rewrite our intuition.

You need to rewrite your intuition. Left cosets are generally not the same as right cosets simply because there is no reason for them to be. When they are the same, that's odd and it means that something interesting is going on. In fact if the two sets of cosets of a subgroup are the same, then that subgroup is special. It is so special that we call it normal! (I suspect that the people who named it that hadn't internalized that noncommutativity is the default, and commuting is surprising. But when nobody has experience with something, it is hard to have good intuition about it.)

So rather than ask why left cosets and right cosets can differ, you should ask why you would have expected them to be the same.

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    $\begingroup$ "In most of our experience, commutativity is standard." I'd say, in most of our experience, commutativity is violated. Putting on your socks, and then your shoes, is not the same as putting on your shoes, then your socks. Buying insurance, then having a fire, is not the same as having a fire, then buying insurance. Even in school mathematics, $3-2\ne2-3$, and $2\div3\ne3\div2$. $\endgroup$ Commented 20 hours ago
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    $\begingroup$ @GerryMyerson What you are saying is true. But most people don't think of "order of physical operations" in a mathematical framework. Similarly it is easy to dismiss subtraction and division as just addition and multiplication by the appropriate inverse. Making the noncommutativity a question of notation, and not what you are doing. $\endgroup$ Commented 17 hours ago
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Let $G = GL_2(\mathbb{R})$ and $H$ be upper triangular matrices:

$$H = \left\{ \begin{pmatrix} a & b \\ 0 & d \end{pmatrix} : a, d \neq 0 \right\}$$

Take $g = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

An element of the Left coset $gH$ has the following form :

$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} a & b \\ 0 & d \end{pmatrix} = \begin{pmatrix} 0 & d \\ a & b \end{pmatrix}$$

An element of the right coset $Hg$ has the following form:

$$\begin{pmatrix} a & b \\ 0 & d \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} b & a \\ d & 0 \end{pmatrix}$$

As sets these are not the same. The key is that the subgroup is not normal.

Contrast this with a normal subgroup. Let $H$ be the subgroup of scalar matrices (multiples of the identity):

$$H = \left\{ \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} : \lambda \neq 0 \right\} = \{\lambda I : \lambda \neq 0\}$$

Take any $g = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in GL_2(\mathbb{R})$

An element of the left coset $Hg$ has the following form:

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} = \lambda \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

An element of the right coset $Hg$ has the following form:

$$\begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \lambda \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

In this case the sets are the same.

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    $\begingroup$ Everything you wrote is correct, but your example of scalar matrices is a tiny bit misleading as that subgroup is central, not just normal, i.e., the $gh = hg$ for all $h \in H$, not just $gH = Hg$. $\endgroup$ Commented 23 hours ago
  • $\begingroup$ This is true, but I assumed the OP knew matrices and this was an easy example to make a point. $\endgroup$ Commented 22 hours ago

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