Question

2. Let G be an abelian group. Suppose that a and b are elements of G of finite order and that the order of a is relatively prime to the order of b. Prove that <a>n<b>= <1> and <a, b> = <ab> .

Answer 1

Since, G be a abelian group then for all x,y belongs to G,

ty = y2

Now, take element a,b of G. Let the order of 'a' is m and order of 'b' is n.

<a> denote the cyclic group which is generated by 'a' that is

<a >= {a', a..., am = a = 11

Then the order of the group will be m and order of each element of <a> divides the order of the group that is 'm'.

Similarly <b> denote the cyclic group which is generated by 'b'that is,

<b>= {61,6..., 6" = b = 1}

Then the order of the group will be n and order of each element of <b> divides the order of the group that is 'n'.

Since,given that gcd(m,n)=1

Hence, common element of the groups <a> and <b> is {1} that is <1>.

Therefore,

<a>n<b>=<1>

Now take (When m,n)

<a, 6 >= {e,a 61, ..., amim = bm ..., a” b” = q”...

Total elements = lcm (m,n)=mn

and

<ab >= {e, (ab),..., (ab) = , ..., (ab)" = a",...}

Total elements = lcm(m,n)=mn

Hence,

<α, ο >=< αν >