Thread: Abstract Algebra

If G is an abelian group that contains elements of order 4 and 10 then what is the order of a*b ?

I know that when the gcd(|a|,|b|)=1 then |ab|=|a||b|, but how do i find the order in this case?

Help will be appreciated.

Hey mrmaaza123.

With regards to Lagrange's theorem we know that a sub-group needs to be a factor of the total order of the group.

Now in the case of gcd being 1, it means that these sub-groups are going to be disjoint (except for the identity element) which means that the order will be the product of the two sub-groups which will have |a|*|b|.

But when you do not have a gcd of one, it means that you need to check the actual sub-groups themselves.

The other alternative is that you factorize the groups into disjoint sub-groups and then apply the situation with gcd being equal to 1.

Thank you for the effort, but i do not know what the group G might be for this question. THe question simply says G.
My book gives the answer as 20, and i am still having some problem in figuring out how that happened.

The factors of both are 2*2 and 2*5. You might want to find theorems that describe decompositions of groups with into ones with only prime factors.

Also this is just an observation but 20 = 4*10/gcd(4,10) = 4*10/2 = 20, so maybe that can give you hint as what to look for.

Aha! i understood that. It seems to work for every question !
I will look up for a proper result though.
But,Thank you very much, i highly appreciate your help.

suppose that |a| = 4, and |b| = 10.

since G is abelian: (ab)² = (ab)(ab) = a(ba)b = a(ab)b = (aa)(bb) = a²b².

an induction proof shows (ab)ⁿ = aⁿbⁿ.

thus (ab)²⁰ = (a)²⁰(b)²⁰ = (a⁴)⁵(b¹⁰)² = e⁵e² = ee = e.

thus |ab| divides 20. since |a| ≠ |b|, b ≠ a^-1 (which has the same order as a), so |ab| ≠ 1.

now (ab)⁴ = a⁴b⁴ = eb⁴ = b⁴ ≠ e, since |b| = 10. so |ab| ≠ 4.

similarly, (ab)¹⁰ = a¹⁰ = a² ≠ e, so |ab| ≠ 10.

thus we are left with: |ab| = 2,5 or 20.

if |ab| = 2, this means a²b² = e, so that a⁴b² = a², that is:

b² = a², and therefore b⁴ = a⁴ = e, contradicting |b| = 10.

if |ab| = 5, then a⁵b⁵ = ab⁵ = e, whence a = b⁵, so that:

a² = b¹⁰ = e, contradicting |a| = 4.

so the only possibility left is |ab| = 20, as no lower divisor of 20 can possibly be the order of |ab|.

in general, for ABELIAN groups (only), if <a>∩<b> = {e}, then indeed, |ab| = lcm(|a|,|b|) in an abelian group (the proof in the general case is pretty much that same:

clearly ab^lcm(|a|,|b|) = e, and if d|lcm(|a|,|b|), with (ab)^d = e,

then a^d = b^|b|-d, so that a^d is in <a>∩<b> and thus equals {e}, so that |a| divides d,

and similarly |b| divides d, so that lcm(|a|,|b|) also divides d, so that d = lcm(|a|,|b|).

the important thing is that <a> and <b> only intersect in {e}, for example:

if b = a^-1, then |ab| = 1, even though lcm(|a|,|a^-1|) = |a|).