group theory: normal subgroups?

Question

i dunno where to start with this question, because i don't really know what it means

Let G be a group. The centre Z(G) of G is defined by

Z(G)={g in G : gx=xg for all x in G}

show that Z(G) is a normal subgroup of G.

Then, let Z=Z(GL(2, reals)), i.e. the set of all 2x2 matricies with real entries and non-zero determinant, that is, Z is the centre of GL(n, reals) and show that Z= the matrix

a 0
0 a

such that a belongs to the set of the reals star, which means all the real numbers except zero.

Any help would be greatly appreciated.

Answer 1

The first part should be easy, pick an element z in Z(G) and g in G and show that g^{-1}zg is in Z(G) (this comes from a theorem about different ways of saying that a subgroup is normal).

For the second one you want to do basically the same thing except you know what the arbritray elements look like and you can do the matrix multiplication.

Answer 2

1). Pick z in Z(G). Let g be any element in G.
Look at g^-1zg. Since z is in the centre of G, this
equals g^-1gz = z. That means g^-1zg is in Z(G)
and we are done.

2. Finally got this one! Let's let A be in the centre of G.
A = a b with a <> 0. The trick is to pick the correct
......c d

elements of G to commute with A.
First, let's try g = 1 1 .
............................1 0
If Ag - gA, we compare the first row, first column
of both products to get
a + b = a + c, i.e., b = c.
So now A has the form
A = a b
.......b d

Now let's choose g = 1 0
.................................... 1 1
and compare Ag and gA,
We get b = 0 and thus c = 0.
Also b + d = a + b,
so a = d
and we are done.
Sorry I can't show every step, but typing
math is a nightmare in this forum!!

Answer 3

each cyclic team is abelian provided that x^nx^m=x^(n+m)=x^(m+n)=x^mx^n. Subgroups of index 2 might desire to have the left and proper cosets equivalent because of the fact cosets are equivalence instructions of equivalent length. So, why are all p-communities solvable?