Proving that a subgroup is normal.

How can we prove that a subgroup H of

Gl_2(Z_3)

is normal?

These are the elements of H:

\begin{pmatrix}1&1\\1&2 \end{pmatrix}

\begin{pmatrix}1&2\\2&2 \end{pmatrix}

\begin{pmatrix}2&1\\1&1 \end{pmatrix}

\begin{pmatrix}2&2\\2&1 \end{pmatrix}

\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}

\begin{pmatrix} -1 & 0 \\0 & -1 \end{pmatrix}

\begin{pmatrix} 0 & 2 \\1& 0 \end{pmatrix}

\begin{pmatrix} 0 & 1 \\2 & 0 \end{pmatrix}

So the determinant of the elements is 1 and the trace is zero.

I don't think I can find the left and right cosets and show that they are equal, because there are too many elements in

Gl_2(Z_3)

. There is a theorem in our textbook that states,

Let H be a subgroup of G. Then H is normal in G if and only if there is a group structure on the set G/H of left cosets of H with the property that the canonical map π : G → G/H is a homomorphism. If H is normal in G, then the group structure on G/H which makes π a homomorphism is unique: we must have gH · gH = ggH for all g, g ∈ G. Moreover, the kernel of π : G → G/H is H. Thus, a subgroup of G is normal if and only if it is the kernel of a homomorphism out of G.

So I'm trying to use this and show that

f: Gl_2(Z_3) \rightarrow Gl_2(Z_3)/H

is a homomorphism...but I'm stuck...

We know that

f(a)f(b)=aHbH

and that

f(ab)=abH

. But how can we show that aHbH=abH? Can anybody please give me a hint?

Thanks in advance

(edited 11 years ago)

Reply 1

ghostwalker

Original post by Artus

...

Perhaps I'm being thick, but what is the definition of H? You say the trace is zero, but some of those matrices do not have a zero trace - the identity element for a start.

Reply 2

Artus

Original post by ghostwalker

Perhaps I'm being thick, but what is the definition of H? You say the trace is zero, but some of those matrices do not have a zero trace - the identity element for a start.

Sorry. H has elements with trace zero and determinant 1 AND additionally the identity element and the -identity element. That's the definition of H.

(edited 11 years ago)

Reply 3

ghostwalker

Can't see any further at the moment. Probably needs someone more familiar with the subject.

Reply 4

Mark85

Original post by Artus

So I'm trying to use this and show that

f: Gl_2(Z_3) \rightarrow Gl_2(Z_3)/H

is a homomorphism...but I'm stuck...

You are going about this ass backwards.

Gl_2(Z_3)/H

is only well defined as a group if

H

is normal. This is the whole point about normal subgroups: Normal subgroups are precisely the kernels of homomorphisms.

Put away the theorems and start thinking about definitions. This is a basic exercise which doesn't require using much theory but rather just understanding the definitions.

If you are given that

H

is a subgroup and need to just prove normality, then you could just show that it is stable under conjugation. That is, for all

g \in GL_2(\mathbb{Z}_3)

and all

h \in H

we have that

ghg^{-1}\in H

.

So, you just need to show/note that the determinant and trace are preserved under conjugation (i.e. similarity) and that in the case of plus/minus the identity (the only cases with trace not equal zero) these are in their own conjugacy classes.

I would have thought that this was immediately apparent and am thus interested in what the actual question was/how it was phrased...

Reply 5

Artus

Original post by Mark85

You are going about this ass backwards.

Gl_2(Z_3)/H

is only well defined as a group if

H

I don't know maybe different textbooks have different definitions, or maybe I'm not understanding it. But when the textbook first introduces G/H for a group G...it does not say anything about normality, it just requires H to be a subgroup. Actually this is my textbook:

http://www.albany.edu/~mark/algebra.pdf (By the way, this pdf document is legal. It's from the author's own webpage.)

It's at the top of the page on p.53 (Definition 3.2.6)

Original post by Mark85

If you are given that

H

is a subgroup and need to just prove normality, then you could just show that it is stable under conjugation. That is, for all

g \in GL_2(\mathbb{Z}_3)

and all

h \in H

we have that

ghg^{-1}\in H

I'll try to think more about this. The reason why I was confused is that there are so many elements in

Gl_2(Z_3)

, so I wasn't sure about how to check wheither

ghg^{-1}

was in H for all g. But again, I'll try to think more about this...

(edited 11 years ago)

Reply 6

Mark85

Original post by Artus

There he is defining a set

G/H

. If you read on, he presumably discusses the fact that the group structure induced by the canonical map is well defined iff H is normal in G.

I would say read more carefully, but frankly, I would advise you to find an easier book.

Original post by Artus

I'll try to think more about this. The reason why I was confused is that there are so many elements in

Gl_2(Z_3)

, so I wasn't sure about how to check wheither

ghg^{-1}

was in H for all g. But again, I'll try to think more about this...

I don't think you understand what I wrote. You don't need to calculate it for each

g

individually. Here it is broken down into smaller pieces for you:

1. Check that H is precisely the set of elements that have trace 0 and determinant 1 together with plus and minus the identity.

2. If you conjugate an element of H by any element of G, it is a basic property of determinants that the resulting element will have determinant 1

3. If you conjugate an element of H (apart from plus minus the identity) by any element of G, it is a basic property of traces that the resulting element will have trace 0.

4. Combining 2. and 3. tells you that if you conjugate an element of H (apart from plus or minus the identity) with any element of G, you get an element with trace zero and determinant 1 which by point 1. is therefore an element of H

5. If you conjugate plus or minus the identity by any element of G you get plus or minus the identity.

6. Putting 4 and 5 together means if you conjugate and element of H with any element of G, you end up back in H.

I seriously recommend you get an easier algebra book - the one you posted is too sophisticated for you. I also recommend you refresh yourself with the basic properties of matrices, determinants and trace.

(edited 11 years ago)

Reply 7

Artus

Original post by Mark85

I seriously recommend you get an easier algebra book - the one you posted is too sophisticated for you. I also recommend you refresh yourself with the basic properties of matrices, determinants and trace.

That's the one we're using in class, so I have to use it. Anyways, thanks about the suggestions, I'll think about them more deeply.

Reply 8

Mark85

Original post by Artus

That's the one we're using in class, so I have to use it. Anyways, thanks about the suggestions, I'll think about them more deeply.

This is nothing deep. This is the very basics of linear algebra.

Matrices are conjugate/similar if they represent the same linear transformation written in different bases. The determinant measures how a linear transformation scales area. Ergo, similar/conjugate matrices must have the same determinant.

The trace of a matrix is the sum of its eigenvalues. Ergo similar/conjugate matrices have the same trace.

Sure, when you use things like determinant and trace it is more convinient to use formulas but if you are thinking about general properties just remember what those things actually are and try and picture them somehow and then it should be easier to at least remember the basic properties.