# Thread: verify the multiplication of two absolute abelian group elements.

1. ## verify the multiplication of two absolute abelian group elements.

**Question**
Given that a and b belongs to an Abelian Group and |a|=6 and |b|=6, whay can you say about |ab|?

**My solution**
|a|=6
a=6
a^6=e

similarly;
|b|=6
b=6
b^6=e

hence
|ab|=(ab)
(ab)=(ab)^6
=a^6b^6
=ee
=e

therefore |ab|=6 and ab belongs to G

Q; Pls what sis I do wrong in answering the question???

Thanks

2. ## Re: verify the multiplication of two absolute abelian group elements.

First, is |a| the order of element a- that is, the power, n, to which a must be raised so that a^n= e? If so what you are doing is correct except where you say "|ab|= (ab)" and (ab)= (ab)^6". I don't know what those could mean. |ab| is, of course, the order of ab but that is a number but what is (ab)? If you just mean "a times b" it is not equal to a number as |ab| is. And I cannot see any interpretation which makes "(ab)= (ab)^6" true. It is true, since G is abelian, that (ab)^6= a^6b^6= 6. To complete the proof that |ab|= 6 you would also need to show that no number less than 6 works. Of course, you do not need to say "and ab belongs to G"- that was given when saying that a and b are both in G.

3. ## Re: verify the multiplication of two absolute abelian group elements.

As you point out, $(ab)^6=e$. So the order of ab is a divisor of 6, namely 1, 2, 3 or 6. Another question arises; is each of these orders actually possible? That is, can you find a group where each possibility is realized? The answer is yes. Hint: look in the direct product of 2 cyclic groups of order 6.

4. ## Re: verify the multiplication of two absolute abelian group elements.

|ab|=?
(ab)^6 =a^6b^6
=ee
=e
hence if
a^n = e
a^6 = e

same applies to b;

where |e| = 1.

hence, (ab)^n = e

as above, (ab)^6 = e..

so what does |ab| stand for then

is |ab| = 1??????