Thread: Group Theory

If x^6 = e where e is the identity element of group G, then is G abelian? If so, give a brief justification, if not, give a counter-example. Thanks.

Originally Posted by bryan06

If x^6 = e where e is the identity element of group G, then is G abelian? If so, give a brief justification, if not, give a counter-example. Thanks.

Did you mean the condition is true for every x in G? Then check S_3

Tonio

thanks for your reply. i did find something online about S_3 but haven't quite got the hang of it yet...would it be possible if you (or anyone who might also be interested) could kindly talk me through the process of completing this question??thanksss

Originally Posted by tonio

Did you mean the condition is true for every x in G? Then check S_3

Tonio

rather than go through a detailed explanation of what S3 is, and how to notate its elements, consider instead the following set:

G = {e,a,a^2,b,ab,a^2b} where multiplication is carried out according to the following rules:

a^3 = b^2 = e, ba = a^2b. note this last rule forces G to be non-abelian (if it is indeed a group).

it should be clear that we can write any product in G as (a^k)(b^m), where k = 0,1, or 2, and b = 0 or 1.

associativity is somewhat tedious to prove, you may take my word for it (one has to examine 432 products in 216 pairs, and verify they are equal,

although you can trim this somewhat by only examining products involving only a's, then only b's, then a mixture of a's and b's-

the rule ba = a^2b means we can always write the a's first).

but it is easy to see we have closure, and a^-1 = a^2, (a^2)^-1 = a, b^-1 = b.

(ab)^2 = a(ba)b = a(a^2b)b = a^3b^2 = ee = e, so ab is its own inverse,

and (a^2b)^2 = a^2(ba)ab = a^2(a^2b)ab = a^4(ba)b = a^4(a^2b)b = a^6b^2 = (a^3)^2b^2 = e^2e = e,

so a^2b is its own inverse, as well. hopefully this will persuade you that G is indeed a group.

so let's raise every element of G to the 6th power:

e^6 = e. a^6 = (a^3)^2 = e^2 = e. (a^2)^6 = a^12 = (a^3)^4 = e^4 = e.

b^6 = (b^2)^3 = e^3 = e. (ab)^6 = ((ab)^2)^3 = e^3 = e.

(a^2b)^6 = ((a^2b)^2)^3 = e^3 = e. so, as you can see, every element of G satisfies x^6 = e,

but G is not abelian, since ba = a^2b ≠ ab.

ahhh i see i see thank you soooo much for your teaching, it's so detailed and very helpful indeed, cheers!

Originally Posted by Deveno

rather than go through a detailed explanation of what S3 is, and how to notate its elements, consider instead the following set:

G = {e,a,a^2,b,ab,a^2b} where multiplication is carried out according to the following rules:

a^3 = b^2 = e, ba = a^2b. note this last rule forces G to be non-abelian (if it is indeed a group).

it should be clear that we can write any product in G as (a^k)(b^m), where k = 0,1, or 2, and b = 0 or 1.

associativity is somewhat tedious to prove, you may take my word for it (one has to examine 432 products in 216 pairs, and verify they are equal,

although you can trim this somewhat by only examining products involving only a's, then only b's, then a mixture of a's and b's-

the rule ba = a^2b means we can always write the a's first).

but it is easy to see we have closure, and a^-1 = a^2, (a^2)^-1 = a, b^-1 = b.

(ab)^2 = a(ba)b = a(a^2b)b = a^3b^2 = ee = e, so ab is its own inverse,

and (a^2b)^2 = a^2(ba)ab = a^2(a^2b)ab = a^4(ba)b = a^4(a^2b)b = a^6b^2 = (a^3)^2b^2 = e^2e = e,

so a^2b is its own inverse, as well. hopefully this will persuade you that G is indeed a group.

so let's raise every element of G to the 6th power:

e^6 = e. a^6 = (a^3)^2 = e^2 = e. (a^2)^6 = a^12 = (a^3)^4 = e^4 = e.

b^6 = (b^2)^3 = e^3 = e. (ab)^6 = ((ab)^2)^3 = e^3 = e.

(a^2b)^6 = ((a^2b)^2)^3 = e^3 = e. so, as you can see, every element of G satisfies x^6 = e,

but G is not abelian, since ba = a^2b ≠ ab.

i want to add here, that if G = <x> (that is, if G consists just of powers of x) and x^6 = e, then G must be abelian.

the reason is: that if we have a,b in G, then a = x^k for some integer k = 0,1,2,3,4,5, and b = x^m for some integer m = 0,1,2,3,4,5

so ab = (x^k)(x^m) = x^(k+m mod 6) = x^(m+k mod 6) = (x^m)(x^k) = ba.

sometimes this is expressed as: powers of x commute (with themselves).

such a group, that is generated by a single element, is called cyclic, and all cyclic groups are abelian.

Originally Posted by Deveno

i want to add here, that if G = <x> (that is, if G consists just of powers of x) and x^6 = e, then G must be abelian.


The second condition is completely superfluous: if G = <x> then G is abelian, whether

x^6 = e is true or not for any, or all, of its elements.

Tonio



the reason is: that if we have a,b in G, then a = x^k for some integer k = 0,1,2,3,4,5, and b = x^m for some integer m = 0,1,2,3,4,5


so ab = (x^k)(x^m) = x^(k+m mod 6) = x^(m+k mod 6) = (x^m)(x^k) = ba.

sometimes this is expressed as: powers of x commute (with themselves).

such a group, that is generated by a single element, is called cyclic, and all cyclic groups are abelian.

.

yes, indeed tonio, i only added it to make the connection with the original post.

i suspect the poster is new to abstract algebra in any form, and may have trouble distinguishing between the statements:

"x^6 = e" and "x is of order 6" (which is not the same thing: the latter implies the former, but NOT vice versa).

in fact, cyclic groups are abelian BECAUSE the integers are abelian (which, in turn, forces the integers modulo n to be abelian).

in a cyclic group, for all intents and purposes, the symbol "x" is just a place-holder, all of the action takes place in the exponents,

which are additive.

anyhow, i suspect the original poster may have actually been asking if a cyclic group of order 6 is necessarily abelian. i base this conjecture

on the nature of the other questions he has asked, and the titles of his threads. in the start of a course on group theory,

one usually has only basic definitions to work with, not any kind of sophisticated overview.

ahh i understand now haha thanksss, it have always thought that x^k=e would imply o(x)=k too but now i will remember it's actually one-way implication

Originally Posted by Deveno

yes, indeed tonio, i only added it to make the connection with the original post.

i suspect the poster is new to abstract algebra in any form, and may have trouble distinguishing between the statements:

"x^6 = e" and "x is of order 6" (which is not the same thing: the latter implies the former, but NOT vice versa).

and yesyes hahaa im new to group theory you're completely right, and thanks so much for your teaching you're amazing!!

Originally Posted by Deveno

in fact, cyclic groups are abelian BECAUSE the integers are abelian (which, in turn, forces the integers modulo n to be abelian).

in a cyclic group, for all intents and purposes, the symbol "x" is just a place-holder, all of the action takes place in the exponents,

which are additive.

anyhow, i suspect the original poster may have actually been asking if a cyclic group of order 6 is necessarily abelian. i base this conjecture

on the nature of the other questions he has asked, and the titles of his threads. in the start of a course on group theory,

one usually has only basic definitions to work with, not any kind of sophisticated overview.