Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    8
    ReputationRep:
    Name:  gym plz.jpg
Views: 198
Size:  28.2 KB

    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

    Name:  dark souls.gif
Views: 152
Size:  764 Bytes

    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?
    Offline

    2
    ReputationRep:
    (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 G \rightarrow G' 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 ghg^{-1} \in H for all g \in G and h \in H

    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 g = \begin{pmatrix} a & 0 \\ c & d \end{pmatrix}, h = \begin{pmatrix} p & 0 \\ p-q & q \end{pmatrix}, and work out what the matrix ghg^{-1} looks like. If the bottom left entry of ghg^{-1} 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.
    • Thread Starter
    Offline

    8
    ReputationRep:
    (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 G \rightarrow G' 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 ghg^{-1} \in H for all g \in G and h \in H

    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 g = \begin{pmatrix} a & 0 \\ c & d \end{pmatrix}, h = \begin{pmatrix} p & 0 \\ p-q & q \end{pmatrix}, and work out what the matrix ghg^{-1} looks like. If the bottom left entry of ghg^{-1} 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 ghg^{-1} matrix I get
    Name:  mattadaq.gif
Views: 116
Size:  596 Bytes

    So this is not normal as the bottom left entry of ghg^{-1} is not equal to the top left entry minus the bottom right entry.

    Is that right?
    Offline

    2
    ReputationRep:
    (Original post by makin)
    Thanks, after doing the ghg^{-1} matrix I get
    Name:  mattadaq.gif
Views: 116
Size:  596 Bytes

    So this is not normal as the bottom left entry of ghg^{-1} is not equal to the top left entry minus the bottom right entry.

    Is that right?
    Reread what I wrote. You made an error in calculating ghg^{-1}.

    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 ghg^{-1} 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.
    • Thread Starter
    Offline

    8
    ReputationRep:
    (Original post by Mark85)
    Reread what I wrote. You made an error in calculating ghg^{-1}.

    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 ghg^{-1} 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 :confused:
    Offline

    2
    ReputationRep:
    (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 :confused:
    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 g:=\begin{pmatrix} 2 & 0 \\ 2 & 2\end{pmatrix} and h:=\begin{pmatrix} 2 & 0 \\ 1 & 1\end{pmatrix} then this wouldn't be a counterexample since by your calculation, you would have

    ghg^{-1} = \begin{pmatrix} 2 & 0 \\ 2 & 1\end{pmatrix} which is clearly in L.

    So, you need to look at your calculation to work out matrices where the conjugate isn't in L.
    • Thread Starter
    Offline

    8
    ReputationRep:
    (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 g:=\begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix} and h:=\begin{pmatrix} 2 & 0 \\ 1 & 1\end{pmatrix} then this wouldn't be a counterexample since by your calculation, you would have

    ghg^{-1} = \begin{pmatrix} 2 & 0 \\ 1 & 1\end{pmatrix} 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
    • Thread Starter
    Offline

    8
    ReputationRep:
    :mad: 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.
    Offline

    2
    ReputationRep:
    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 H 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 pq \neq 0 \neq ad and

    \frac{cp + dp -dq - qc}{a} \neq {p-q}

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

    \frac{cp + dp -dq - qc}{a}= (p-q)\frac{(c+d)}{a}

    Therefore, any number c together with any nonzero numbers p,q,a,d such that p\neq q and c+d \neq a 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!)
    • Thread Starter
    Offline

    8
    ReputationRep:
    (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 H 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 pq \neq 0 \neq ad and

    \frac{cp + dp -dq - qc}{a} \neq {p-q}

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

    \frac{cp + dp -dq - qc}{a}= (p-q)\frac{(c+d)}{a}

    Therefore, any number c together with any nonzero numbers p,q,a,d such that p\neq q and c+d \neq a 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
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.