Example of Group Action

Matrix Theory: Consider the set G of matrices of the form [x y \ 1 0] where x is nonzero real and y is real. Let G act on the real line R by [x y \ 1 0].t = xt + y. Show that G is a group, that the action is a group action, and that the action is faithful.

[michaelraum]

I seriously love your videos, always have.

[MathDoctorBob]

@cpaniaguam That is exactly the idea. The action defines a homomorphism of the group into the symmetry group of the set. Faithful means one-one, so we are identifying the group with a subgroup of symmetries. Otherwise we have some quotient of the group as a subgroup of symmetries.

Worth the reminder: each g in the group acts the set as a one-one correspondence. - Bob

[MathDoctorBob]

:) It's actually good form on your part. You don't truly understand definitions or theorems until you stress test them with examples, especially ones where they fail.

[profezmo]

O wow I knew I was missing something obvious, you truly are the math doctor

[MathDoctorBob]

Checking Wolfram, they are the same. I haven't seen "free" used as "faithful, " but not surprising. Both mean the only group element that fixes everything is the identity.

[profezmo]

Bit of confusion, say we have the group GL2(R) acting on the set R^2. If we use the definition of a faithful group action that says g(x) = x is only true when g=e then that means this IS a faithful group action.

But it doesn't appear to be one-to-one: say we have g = {1 1; 1 1} and s = {2 4} and t = {4 2}, then g(s) = {6 6} and g(t) = {6 6}

So g(s) = g(t) but s does not equal t which would mean it's not one-to-one. What am I doing wrong here?

Thanks for the video btw

[MathDoctorBob]

@YesITMarketing You're welcome! - Bob

[YesITMarketing]

Thank you! You've done a good job!

[profezmo]

Another question, what exactly is the difference between a free group action and a faithful group action? Because according to wolfram, they seem to be the same thing...

[cpaniaguam]

So when a group acts on a set faithfully means that the action is a one to one correspondence? To me this is reminiscent of kernels of group homomorphisms f for which f^{-1}(e')={e}.

[MathDoctorBob]

You're welcome!

det [1 1\ 1 1] = 0 so it is not invertible and not one-one as a map from R2 to itself. Note that it is not in GL(2, R).

[MathDoctorBob]

@dfnmartins Thanks! I'd love to visit Brazil. No sticks for dummies, just for pointing. - Bob

[categorygrp]

I’m confused about what you’re doing at 5:40 with applying a matrix to a scalar to get a scalar equation. How does that work or where can I go to learn what you’re doing?