Automorphisms over a quotient group of GLn\SLn over Z7

**adam63** · August 18th 2010, 01:41 AM

I have a problem that I have no idea how to begin solving. I need to describe the following group using an 'action table' (when the action is composition of homomorphisms). Moreover, I need to find out what is isomorphic to this group.

The group is : $Aut(GL_n(Z_7) / SL_n(Z_7))$ , which is a quotient group, with a given n>0, and knowing that $Z_7$ is a field.

Now, I believe I need to find $GL_n(Z_7) , SL_n(Z_7)$ , and find the order of each group. Then, I'll be able to know how many cosets the quotient group contains. Then, I'll have to actually FIND and describe the different automorphisms on this quotient group... Afterward, I'll need to check what happens whenever I composite each and every automorphism to another, and put it in a table, as I was asked to.

At last, I'll need to find what this group of automorphisms is isomorphic to...

Any help

??

Thanks a million!

**tonio** · August 24th 2010, 06:34 PM

Originally Posted by adam63

I have a problem that I have no idea how to begin solving. I need to describe the following group using an 'action table' (when the action is composition of homomorphisms). Moreover, I need to find out what is isomorphic to this group.

The group is : $Aut(GL_n(Z_7) / SL_n(Z_7))$ , which is a quotient group, with a given n>0, and knowing that $Z_7$ is a field.

Now, I believe I need to find $GL_n(Z_7) , SL_n(Z_7)$ , and find the order of each group. Then, I'll be able to know how many cosets the quotient group contains. Then, I'll have to actually FIND and describe the different automorphisms on this quotient group... Afterward, I'll need to check what happens whenever I composite each and every automorphism to another, and put it in a table, as I was asked to.

At last, I'll need to find what this group of automorphisms is isomorphic to...

Any help

??

Thanks a million!

I'd rather write $\mathbb{F}_7:=\mathbb{Z}_7=\mathbb{Z}/7\mathbb{Z}$

Hints:

(1) For $q=p^r\,,\,\,p$ a prime, we have that $|GL_n(\mathbb{F}_q)|=(q^n-1)(q^n-q)\cdot\ldots\cdot(q^n-q^{n-1})=q^{n(n-1)/2}\prod\limits_{i=1}^n(q^i-1)$ . Hint for proof: count how many inversible matrices are there by counting in how many ways can you choose the first column of such a matrix, then in how many ways can its 2nd column be chosen, etc.

2) Since $\Delta:GL_n(\mathbb{F}_q)\rightarrow \mathbb{F}_q^{*}\,,\,\,\Delta(A):=\det(A)$ is an epimorphism of groups, we get that $|SL_n(\mathbb{F}_q)|=\frac{|GL_n(\mathbb{F}_q)|}{q-1}$ (Hint: what is $\ker(\Delta)$ ?)

3) From the above it follows at once that $Aut\left(GL_n(\mathbb{F}_q)/SL_n(\mathbb{F}_q)\right)$ is the automorphism group of a cyclic group, so...

Tonio

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