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Normal Subgroup Questions

  1. May 3, 2007 #1
    Having trouble with a couple of algebra questions and would really appreciate any hints or pointers.

    1. A is a subgroup of group G with a finite index. Show that
    [tex]N = \bigcap_{x \in G}x^{-1}Ax[/tex]
    is a normal subgroup of finite index in G.

    I'm able to show that N is a subgroup of G by applying the subgroup test. Thing is, I'm not sure how to prove that it's a normal subgroup. It seems that the fact that A is of finite index should play into it somehow.

    2. Let [tex]G = GL(n,\mathbb{Z})[/tex] for [tex]n \ge 2[/tex]. Define the n-th converge subgroup, G(m), as [tex]G(m) = \left\{A \in G : A\equiv I_n\mod m\right\}[/tex].
    Show that G(m) is a normal subgroup.

    Tried thinking of this as x^-1yx where x is just a GL matrix and y is one of G(m) and trying to show that this product is one of G(m). Wrote some formulas for individual entries of the product matrix, but doesn't seem to work in terms of guaranteeing that each non-diagonal entry is a multiple of m, and every digonal entry is a multiple of m and with an extra 1. Although perhaps it's just that this gets somewhat messy and I made some silly mistake somewhere.

    Thanks in advance. Any help is really appreciated.
    Last edited: May 3, 2007
  2. jcsd
  3. May 3, 2007 #2

    matt grime

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    1. if y is in N, is x^-1yx?

    2. It seems clear to me that (AB) mod m (I presume m is an integer, and mod m means take the entries mod m) is equal to (A mod m)(B mod m) mod m, since multiplication is just addition and multiplication of the entires of A and B. So normality is straightforward in 2. I think your method does work, though it is messy. If you want another way to think about it, A=I mod m means that A=I+A', and A' is a matrix where every entry is divisible by m. That should be easier to visualize, and for you to prove it directly. You should at least show that the set of matrices {B:B=0 mod m} is an ideal.
  4. May 3, 2007 #3
    Thanks for your help. The hint for the first one was a good step in the right direction.

    One the second one, I've made progress, but I could use a bit of clarification on two points.
    -First, to show that G(m) is a subgroup of G one of the things I need to do is show that if A is in G(m) then its inverse is also in G(m). I'm having trouble proving that point.
    -Secondly, I tried going about proving normality in the most conventional manner showing that gG(m)=G(m)g (for all g belonging to G). In other terms, for all g belonging to G and for all a belonging to G(m), ga=ag. Thus, g(a'+I)=(a'+I)g. This simplifies to ga'=a'g, but this isn't true in general for matrices. Where am I going wrong?
  5. May 3, 2007 #4

    matt grime

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    You're going wrong becuase you're not reducing mod m. You're not supposed to show what you're trying to show - remember the group G(m) is, or ought to be, with multiplication mod m too.
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