Question

Prove the converse of lagrange's theorem hold for abelian groups

Answer 1

Let

G

G

a finite group. If

a∈G

a∈G

, a consequence of the Lagrange's Theorem is that the order of

a

a

divides the order of

G

G

. Let

p=|G|

p=|G|

. If

p

p

is prime, it is well known that

G

G

is cyclic, and therefore, there exists

a∈G

a∈G

such that the order of

a

a

is

p

p

and, of course, exist

b

b

in

G

G

such that the order of

b

b

is

1

1

. If

n

n

is a composite number, namely

n=

p

m

1

1

p

m

2

2

…

p

m

n

n

,

n=p1m1p2m2…pnmn,

where

p

i

pi

are primes, my question is this:

Is there, for each

i

i

, an element

a

i

ai

in

G

G

such that the order of

a

i

ai

is

p

i

pi

? Is there, for each

i

i

,

b

i

∈G

bi∈G

, such that the order of

b

i

bi

is

p

m

i

i

pimi