DAILY MATH CHALLENGE THREAD #3

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>>8670377
And here is #2.

>>8670377
i don't know any group theory

>>8670380
That's why there's a 2nd problem.

>>8670379
Hey, OP. Any tips for this one?

>>8670377
my group theory is so rusty, all I know is that there's a nontrivial a s.t. a^2 = e and I've been trying to work with that. any hints?

>>8670379
9/10

>>8670377
can you just take the kernel of the map f: G to G defined by g to g^k? since f^2=0 map

Give a pair of infinite fields where the additive group of one is isomorphic to the multiplicative group of the other.

>>8670453
Telescoping series.

>>8670507
Let the mapping φ: G --> S_G be the regular representation. Show that if φ(x) has an odd permutation then G has a subgroup of index 2.

>>8670522
Correct!

>>8670538
Not sure I follow, can you explain?

>>8670538
it's possible that f is the 0 map, say if the group is [math]\mathbb{Z}_2 \times /mathbb{Z}_k[\math]. In that case the kernel of f is everything

>>8670566
I mean [math]\mathbb{Z}_2 \times \mathbb{Z}_k[/math]

>>8670453
[math] \dfrac {1}{a+b} = \dfrac {1}{a+b}\dfrac {a-b}{a-b}=\dfrac {a-b}{a^2-b^2} [/math]
Similar to what you do with complex numbers.

>>8670570
why does g^k kill Z_2 if k is odd?

>>8670592
nevermind, I was being dumb. You are right about Z_2.

Still, how do we know that the kernel has k elements? Also (and maybe this is a dumb question) why is the kernel even a subgroup? If G is nonabelian I don't see why xy must be in the kernel if x and y both are.

>>8670605
>Also (and maybe this is a dumb question) why is the kernel even a subgroup?
kernels are always subgroups (in fact they're exactly the normal subgroups of a given group), and you need to check xy^{-1} for the subgroup test, not xy

>>8670621
The kernel of a homomorphism is always a subgroup. Why is your f a homomorphism?