Thread: Group theory questions

I did a test recently and I struggled on these questions

(1) Prove in any group (x^-1)^-1=x

I said (x^-1)^-1=a so a^-1(x^-1)^-1=e

so (x^-1*a)^-1=e implying x^-1*a=e, implying a=x.

But that's assuming what is to be proved. namely that the inverse of x^-1 is x.
(2) Prove if a and b are conjugate, they have the same order.

My answer. Let order of a=k and let e be the identity.
a=gbg^-1 for some g
a^k=(gbg^-1)^k=(g^-1)^k*b^k*g^k=e
This implies b^k=e so order of b=k. But I don't think that's fully correct because it doesn't show k is the smallest integer such that b^k=e.

(3) decide which of the following are isomorhic, quoting results you use

Z2 xZ6, Z3XZ4, Z2 X Z2 X Z3, Z12. My failure is just ignorance with respect to these standard results. I only knew that Z12 is cyclic. For Zm X Zn to be cyclic, m and n need to be co-prime. 2 and 6 are not co-prime so Z2 X Z6 is not cyclic and so not isomorphic to Z12.

Help much appreciated.

Originally Posted by Duke

I did a test recently and I struggled on these questions

(1) Prove in any group (x^-1)^-1=x

I said (x^-1)^-1=a so a^-1(x^-1)^-1=e

so (x^-1*a)^-1=e implying x^-1*a=e, implying a=x.

But that's assuming what is to be proved. namely that the inverse of x^-1 is x.

Starting from , multiply both sides (on the left) by . Then multiply both sides (on the left) by x.

Originally Posted by Duke

(2) Prove if a and b are conjugate, they have the same order.

My answer. Let order of a=k and let e be the identity.
a=gbg^-1 for some g
a^k=(gbg^-1)^k=(g^-1)^k*b^k*g^k=e
This implies b^k=e so order of b=k. But I don't think that's fully correct because it doesn't show k is the smallest integer such that b^k=e.

What that argument shows is that (order of b) ≤ (order of a). But if b is conjugate to a, then a is conjugate to b. So you can switch a and b in that argument, getting the reverse inequality.

Originally Posted by Duke

(3) decide which of the following are isomorphic, quoting results you use

Z2 xZ6, Z3XZ4, Z2 X Z2 X Z3, Z12. My failure is just ignorance with respect to these standard results. I only knew that Z12 is cyclic. For Zm X Zn to be cyclic, m and n need to be co-prime. 2 and 6 are not co-prime so Z2 X Z6 is not cyclic and so not isomorphic to Z12.

It's probably best to start by looking at the orders of elements in these groups. is the only one to have an element of order 12, so that sets it apart from any of the other groups. In each of the remaining groups, is there an element of order 4? How many elements have order 6? By looking at questions like that, you can determine that some of the groups are nonisomorphic to others. If there are any groups that you cannot distinguish by properties like that, then you will have to think about whether they are in fact isomorphic.

(1)x^-1(x^-1)^-1=x^-1a

implying x^-1a=e

x*x^-1*a=x

What now?

(2) Ok I get it thanks.

(3)what about Z3 XZ4. Surely the order of (2,3) =lcm of 3 and 4 =12

Originally Posted by Duke

(3)what about Z3 XZ4. Surely the order of (2,3) =lcm of 3 and 4 =12

You're right, of course. My mistake. 3 and 4 are co-prime, so is cyclic of order 12 and therefore isomorphic to .

what about the first one?

Originally Posted by Duke

what about the first one?

You may find it helpful that

sorry I meant my first question, (x^-1)^-1=x. I feel rather stupid but I still can't prove it.

well from the definition of inverse:

(x^-1)(x^-1)^-1 = e. therefore:

x[(x^-1)(x^-1)^-1] = xe = x. and by associativity:

(x(x^-1))(x^-1)^-1 = x

e(x^-1)^-1 = x

(x^-1)^-1 = x. no extra symbol "a" required.

Ah, I got confused. I forgot I still can use the fact that xx^-1=e. Hang on surely it follows straight away that (x^-1)^-1=x.

yes, it is a straight-forward consequence of what it means to be an inverse. this is the origin of the grade school maxim: "the negative of a negative is positive", or

"the recriprocal of 1/a is a".

Yeah a bit strange asking me to prove it.

proving (x^-1)^-1 = x, and (ab)^-1 = b^-1a^-1 are standard proofs often asked for in a first course in groups.

they aren't "hard" questions, they are just designed to see if you "get it".

How strange you mention that. That proof was asked alongside it and could do that as it's not so much of an axiom as the other.