# Algebra, element orders

• May 4th 2008, 01:28 AM
Kai
Algebra, element orders
Hi, need quick help with these questions, have exams on Tuesday :(

1) Let G be a group. Prove that
$(xax^-1)^n=xa^nx^-1$

I have been able to prove it, first with n=1, then assuming it holds for n=k and proving for n=k+1 and n=-m for some +ve int m

But the lecturer used n=0 for the first one,

For n=0, he got $(xax^-1)^0=e$ and for RHS $xa^0x^-1=xex^-1$

I dont understand how $a^0=e$ or $(xax^-1)^0=e$

2) Let G be a group and a is an element of G. Prove $o(a)=o(a^-1)$

For second question i have no idea at all, what is o(a) ??

Third question i have been able to do, needs to verify if its corect

3) A relation R is defined on Z by aRb iff a-b is divisible either by 5 or by 7(a,b in Z). Is R an equivalence relation on Z? Justify your answer.

For reflexive, 5|a-a or 7|a-a = 5|0 or 7|0
Hence aRa.

For symmetric, aRb implies 5|a-b or 7|a-b = 5|b-a or 7|b-a which implies bRa.

For transitivity, aRb and bRc implies
5|a-b or 7|a-b and 5|b-c or 7|b-c
= 5|a-b+b-c or 7|a-b+b-c
= 5|a-c or 7|a-c
=aRc

Hence R is an equivalence relation on Z

• May 4th 2008, 03:47 AM
Unenlightened
Quote:

Originally Posted by Kai
Hi, need quick help with these questions, have exams on Tuesday :(

1) Let G be a group. Prove that
$(xax^-1)^n=xa^nx^-1$

I have been able to prove it, first with n=1, then assuming it holds for n=k and proving for n=k+1 and n=-m for some +ve int m

But the lecturer used n=0 for the first one,

For n=0, he got $(xax^-1)^0=e$ and for RHS $xa^0x^-1=xex^-1$

I dont understand how $a^0=e$ or $(xax^-1)^0=e$

I haven't done groups in ages, but as far as I recall we used to use e as to symbolise the identity, which for multiplication would be 1, which would make sense for your a^0 = 1.

For the second question o(a) is the order of a...
And o(a) = smallest m such that a^m = e.

• May 4th 2008, 04:22 AM
Kai
Quote:

Originally Posted by Unenlightened
I haven't done groups in ages, but as far as I recall we used to use e as to symbolise the identity, which for multiplication would be 1, which would make sense for your a^0 = 1.

Yes under multiplication it would make sense, but the binary operation on G has not been defined. it could be addition or any other type
• May 4th 2008, 04:56 AM
Isomorphism
Quote:

Originally Posted by Kai
Hi, need quick help with these questions, have exams on Tuesday :(

1) Let G be a group. Prove that
$(xax^-1)^n=xa^nx^-1$

I have been able to prove it, first with n=1, then assuming it holds for n=k and proving for n=k+1 and n=-m for some +ve int m

But the lecturer used n=0 for the first one,

For n=0, he got $(xax^-1)^0=e$ and for RHS $xa^0x^-1=xex^-1$

I dont understand how $a^0=e$ or $(xax^-1)^0=e$

Thats a debatable topic. Some may call it a convenient notation.
Convenient because:

$a\cdot a^{-1} = e \text{ and } a\cdot a^{-1} = a^{(1-1)} = a^0$

Quote:

Originally Posted by Kai
2) Let G be a group and a is an element of G. Prove $o(a)=o(a^-1)$

For second question i have no idea at all, what is o(a) ??

o(a) is the order of a.

The order of an element a in a group is the least positive integer k such that
$a^k$ is the identity.

Quote:

Originally Posted by Kai
Third question i have been able to do, needs to verify if its corect

3) A relation R is defined on Z by aRb iff a-b is divisible either by 5 or by 7(a,b in Z). Is R an equivalence relation on Z? Justify your answer.

For reflexive, 5|a-a or 7|a-a = 5|0 or 7|0
Hence aRa.(Clapping)

For symmetric, aRb implies 5|a-b or 7|a-b => 5|b-a or 7|b-a which implies bRa.(Clapping)

For transitivity, aRb and bRc implies
5|a-b or 7|a-b and 5|b-c or 7|b-c
Isomorphism: What if 5|a-b and 7|b-c?

• May 4th 2008, 05:51 AM
Kai
But if u see it this way, then for reflexive, 5|0 AND 7|0 not 5|0 or 7|0