Some more Group-theory proofs/questions

[daon]

1) Not sure if this is correct: Suppose a group contains element a and b such that |a| = 4 and |b| = 2 and a³b = ba. Find |ab|. (note: |x| denotes the order of element x.)

I reasoned that since a⁴ = e, then (aa)(aa) = e or aa = a^-1a^-1. Similarly, I got b^-1 = b.

So, a³b = ba <=> aaab = ba <=> aaab^-1 = ba (since b is its own inverse)
<=> aaab^-1b = bab <=> aaa = bab <=> aaaa = abab.

We know the LHS is e, so e = abab or e = (ab)². So |ab| = 2.

2) I am severely stuck on this one: Let G be a group. Show that Z(G) = \(\displaystyle \L \cap_{a \in G}\)C(a). I was thinking that I should try and show contaiment both ways. I.e. that if x is in Z(G) then x is in the intersection of the centralizers of a in G, and vice versa. However, I am confused with the definitions of C(G), C(g) and Z(G).

3) General question here: If I have a matrix and am asked to find the order of it, I am looking for the power of A that is the identity matrix, correct? And, if this matrix never returns to the identity, it is of infinite order?

4) I have a couple questions about this one: Let G be a group of fns from R to R* with the operation of function multiplication. Let H = {f \(\displaystyle \in\) G | f(2) = 1}. Prove that H is a subgroup of G.

- What does the * stand for in this context? From CS experience, I think of the star operation as the collection of all finite substrings taken from an alphabet or language. In have never seen this operation in number theory.
- This question confuses me a bunch. What would the identity elt be, and how can I find an inverse? Is this enough to show closure: given f₁ and f ₂ \(\displaystyle \in\)G f₁(2)*f₂(2)=1*1=1.

Thanks for your help in advance.

-Daon

 

[daon]

OK, I think I figured out #2. I just kept rereading the definitions until I felt sick, then it made a little more sense

(forward)
If x\(\displaystyle \in\)Z(G) then xa=ax for all a\(\displaystyle \in\)G. That is, x commutes with all elts in G.
=> x\(\displaystyle \in\)C(a) for all a\(\displaystyle \in\)G
=> x\(\displaystyle \in\) intersection of C(a) for all a.

(backwards)
If x\(\displaystyle \in\) intersection of C(a) for all a\(\displaystyle \in\) thenx\(\displaystyle \in\)C(a) for all a\(\displaystyle \in\)G. That is, x commutes with all elements in G.
=> x\(\displaystyle \in\)Z(G).

I hope this is right. The second seems almost the exact reverse of the first.

Still struggling with The last one I posted...
-Daon

 

[daon]

For the last one that I am having trouble with, I thought a little further:

I am asked to prove that it is a subgroup, but I can't find an inverse or identity. If I let f be in H, then f(2)=1. But then how can f^-1(2)=1 unless f=f^-1? What is required to show that the inverse is in H and that thay are equal? Then is the identity any function in H? Because for all f,g in H f*g=1=f=g?

Thanks

 

[pka]

To show that a subset of a group is a subgroup it sufficient to show that the set is closed with respect to group operation and group inverse.
You have shown operational closure.
Now for inverse: if j belongs to H then j belongs to G.
We are given that G is a group so j<SUP>-1</SUP> exits in G.
If e is the identity in G then j<SUP>-1</SUP>(x)j(x)=e(x)=j(x)j<SUP>-1</SUP>(x), j<SUP>-1</SUP>(1)=2. We need j<SUP>-1</SUP>(2)=1.
But j<SUP>-1</SUP>(2)j(2)=e(2)=1 or j<SUP>-1</SUP>(2)=1.

 

[daon]

pka, The operation is not composition, but multiplication. But your example has helped me figure it out (I think.. assuming 1 is the identity elt):

If f \(\displaystyle \in\) H, then we know f^-1 \(\displaystyle \in\) G. So, f(2)=1 <=> f(2)*f^-1(2) = 1*f^-1(2) <=> 1 = f^-1(2).

- Is 1 the identity elt?

 

[pka]

See my correction above.

 

[daon]

Many thanks