# Showing that a certain subgroup is the kernal of a homomorphism

1. Oct 4, 2011

### demonelite123

Let Q be the quaternion group {1, -1, i, -i, j, -j, k, -k}. Show that the normal subgroup {1, -1, i, -i} is the kernal of a homomorphism from Q to {1, -1}.

I know that if N is a normal subgroup of G then the homomorphism f: G -> G/N has N as the kernal of f. while i can get the kernal of f to be {1, -1, i, -i} i can't seem to get the codomain to be {1, -1}. i've gone through different mappings and all of them while they had the correct kernal always seem to have more than just {1, -1} in the codomain. can someone help me determine the correct homomorphism?

2. Oct 5, 2011

### Hurkyl

Staff Emeritus
Can you show your work? Why do you think the codomain has more than 2 elements?

Every element of Q appears in the image. The point is that they aren't all distinct in the image.

3. Oct 7, 2011

### demonelite123

i have tried mapping x to x^2 but then i2 = -1 so it is not in the kernal, x^4 would exclude -1 in the set {1, -1}. i have also tried mapping x to xN where N = {1, -1, i, -i} is a normal subgroup so there exists a natural homomorphism with N as its kernal. but that doesn't work since for instance when x = k we would get kN = {k, -k, -j, j} which is not the right codomain.

would this next map work? $x \mapsto xN$ where N = {1, -1, i. -i}, x is any member of Q, xN is a left coset, and N is a normal subgroup of Q. Then i map again: $xN \mapsto (xN)^2$. so the composition of these will map an x to (xN)2 and the codomain will be {1, -1}.

4. Oct 7, 2011

### Deveno

no because the map x→(xN)2 maps everything to the trivial group {N}, so its kernel is Q.

you were on the right track mapping x→xN, can you think of a way to create an isomorphism of Q/N with {-1,1}?

5. Oct 10, 2011

### demonelite123

ok so i know i can't square it since i would just get x2N = N which would be the identity. i was thinking earlier $$xN \mapsto x(xN)$$ but that just gives me N back again. what i want to do is to "kinda" square all the entries somehow in xN so that i get only 1 and -1 but i cannot figure out a mapping to help me do that.

6. Oct 11, 2011

### demonelite123

how about $xN \mapsto x^2$? that way {1, -1, i, -i} = N maps to 1 and {j, -j, k, -k} = (k)^2 = -1

7. Oct 11, 2011

### Deveno

the map xN → x2 is not well-defined.

for x = i, we get N→i2 = -1
for x = 1. we get N→12 = 1

so we do not have a unique image for the coset N.

8. Oct 13, 2011

### demonelite123

hm i'm really stumped on this. it's probably not that difficult but i just don't see it. can someone offer a hint or two?

9. Oct 14, 2011

### Deveno

if N = {1,-1,i,-i} then φ:Q→Q/N given by x→xN, has kernel N.

since Q/N and {-1,1} are both cyclic groups of order 2, there is an isomorphism between them, namely:

N→1
jN(= kN)→-1, if we call this isomorphism ψ, then ψφ is the desired homomorphism.

10. Oct 14, 2011

### demonelite123

ah i didn't notice that they were cyclic groups! i was trying to come up with an explicit mapping between them (like a formula of some sort) but could not come up with one at all. this is my first time being exposed to these kind of ideas and i think i'm slowly starting to understand more of it. thanks for your help!