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# Normal subgroup of a matrix watch

1. Hi everyone. I'm trying to see if the subgroup H, is a normal subgroup. After conjugating H with the general elements of L, I get the below 2x2 matrix

I'm having a hard time deciding whether this is normal or not.
Is it not a normal subgroup because it does not have equal diagonal entries to the group L?
2. (Original post by makin)
Is it not a normal subgroup because it does not have equal diagonal entries to the group L?
I don't even know what you mean by this. You seem to jump back and forth between talking about single matrices and groups of matrices.

To answer the question, you simply have to look at the definition of a normal subgroup again.

The best definition (although normally not the first one given in books that I have read) of a normal subgroup is that it is the kernel of a homomorphism. Therefore, to show a subgroup H of a group G is normal, you can show that it is the kernel of some homomorphism from G to another group G'

However, another way to show a subgroup H of G is normal (and this is normally the first definition) is to show that it is stable under conjugation; more precisely that for all and

So, all you do is take an arbitrary element of H, conjugate by an arbitrary element of G and see what you get. If it is an element of H then H is normal. It looks like you tried to do the first part, although I think you made a slight error with your matrix multiplication. In any case, to show that H isn't normal, you could do the same thing - conjugate an arbitrary element of H by an arbitrary element of G. If what you get isn't generally in H then you just need to pick one example to show that H isn't normal.

So in this case, you take , , and work out what the matrix looks like. If the bottom left entry of is equal to the top left entry minus the bottom right entry, then clearly it is in H and so H is normal. Otherwise, if this isn't generally the case - you use the formula to pick simple values of a,c,d and p,q to find a matrix h in H and a matrix g in L such that when you conjugate h by g - you end up outside H which is enough to show that H isn't normal.
3. (Original post by Mark85)
I don't even know what you mean by this. You seem to jump back and forth between talking about single matrices and groups of matrices.

To answer the question, you simply have to look at the definition of a normal subgroup again.

The best definition (although normally not the first one given in books that I have read) of a normal subgroup is that it is the kernel of a homomorphism. Therefore, to show a subgroup H of a group G is normal, you can show that it is the kernel of some homomorphism from G to another group G'

However, another way to show a subgroup H of G is normal (and this is normally the first definition) is to show that it is stable under conjugation; more precisely that for all and

So, all you do is take an arbitrary element of H, conjugate by an arbitrary element of G and see what you get. If it is an element of H then H is normal. It looks like you tried to do the first part, although I think you made a slight error with your matrix multiplication. In any case, to show that H isn't normal, you could do the same thing - conjugate an arbitrary element of H by an arbitrary element of G. If what you get isn't generally in H then you just need to pick one example to show that H isn't normal.

So in this case, you take , , and work out what the matrix looks like. If the bottom left entry of is equal to the top left entry minus the bottom right entry, then clearly it is in H and so H is normal. Otherwise, if this isn't generally the case - you use the formula to pick simple values of a,c,d and p,q to find a matrix h in H and a matrix g in L such that when you conjugate h by g - you end up outside H which is enough to show that H isn't normal.

Thanks, after doing the matrix I get

So this is not normal as the bottom left entry of is not equal to the top left entry minus the bottom right entry.

Is that right?
4. (Original post by makin)
Thanks, after doing the matrix I get

So this is not normal as the bottom left entry of is not equal to the top left entry minus the bottom right entry.

Is that right?

Also, in any case, even if what you wrote were correct - I think that to fully justify your answer here you should demonstrate more thoroughly that isn't necessarily in H by picking some numbers for a,c,d and p,q (i.e. pick an actual matrix in H and an actual matrix in L) and show that the conjugate is something clearly not in H.
5. (Original post by Mark85)

Also, in any case, even if what you wrote were correct - I think that to fully justify your answer here you should demonstrate more thoroughly that isn't necessarily in H by picking some numbers for a,c,d and p,q (i.e. pick an actual matrix in H and an actual matrix in L) and show that the conjugate is something clearly not in H.
I can't see where I've gone wrong in my calculations as I checked earlier, when you initially mentioned it. Plus I've checked again but cannot see where I have gone wrong
6. (Original post by makin)
I can't see where I've gone wrong in my calculations as I checked earlier, when you initially mentioned it. Plus I've checked again but cannot see where I have gone wrong
Sorry, I misread it - it is correct. However, the rest of the comment remains valid.

I wouldn't have thought it would be enough to claim that the general element you find isn't in H, I think you should really choose values of the parameters and exhibit an explicit matrix g in L and h in H such that the conjugate of h by g isn't in H.

For instance, if you chose and then this wouldn't be a counterexample since by your calculation, you would have

which is clearly in L.

So, you need to look at your calculation to work out matrices where the conjugate isn't in L.
7. (Original post by Mark85)
Sorry, I misread it - it is correct. However, the rest of the comment remains valid.

I wouldn't have thought it would be enough to claim that the general element you find isn't in H, I think you should really choose values of the parameters and exhibit an explicit matrix g in L and h in H such that the conjugate of h by g isn't in H.

For instance, if you chose and then this wouldn't be a counterexample since by your calculation, you would have

which is clearly in L.

So, you need to look at your calculation to work out matrices where the conjugate isn't in L.
I'll try some values to try and get it so it is not in L. I really appreciate the help on this so thank you
8. Still can't find the values. Is there a method to picking them as I have been using a combination of zeroes and ones for both.
9. There is no method - it is just basic arithmetic but you need to think about the situation and not make arbitrary restrictions...

...Why did you try and use ones and zeros? If you think about it for a second, the only element of that consists entirely of ones and zeros is the identity matrix and that is clearly fixed under conjugacy by any element.

You need to find p,q and a,c,d such that and

It might be better to write this in a more natural form e.g.

Therefore, any number c together with any nonzero numbers p,q,a,d such that and are going to work.

Note that I made a mistake in my previous post (now updated) and what I said wasn't a counterexample in fact was (too late!)
10. (Original post by Mark85)
There is no method - it is just basic arithmetic but you need to think about the situation and not make arbitrary restrictions...

...Why did you try and use ones and zeros? If you think about it for a second, the only element of that consists entirely of ones and zeros is the identity matrix and that is clearly fixed under conjugacy by any element.

You need to find p,q and a,c,d such that and

It might be better to write this in a more natural form e.g.

Therefore, any number c together with any nonzero numbers p,q,a,d such that and are going to work.

Note that I made a mistake in my previous post (now updated) and what I said wasn't a counterexample in fact was (too late!)
Thanks again Mark x

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