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.
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.
suppose that |a| = 4, and |b| = 10.
since G is abelian: (ab)^{2} = (ab)(ab) = a(ba)b = a(ab)b = (aa)(bb) = a^{2}b^{2}.
an induction proof shows (ab)^{n} = a^{n}b^{n}.
thus (ab)^{20} = (a)^{20}(b)^{20} = (a^{4})^{5}(b^{10})^{2} = e^{5}e^{2} = 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} = a^{4}b^{4} = eb^{4} = b^{4} ≠ e, since |b| = 10. so |ab| ≠ 4.
similarly, (ab)^{10} = a^{10} = a^{2} ≠ e, so |ab| ≠ 10.
thus we are left with: |ab| = 2,5 or 20.
if |ab| = 2, this means a^{2}b^{2} = e, so that a^{4}b^{2} = a^{2}, that is:
b^{2} = a^{2}, and therefore b^{4} = a^{4} = e, contradicting |b| = 10.
if |ab| = 5, then a^{5}b^{5} = ab^{5} = e, whence a = b^{5}, so that:
a^{2} = b^{10} = 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|).