Exercise 10.D.2 from Pinter says:
Let
a
be any element of finite order of a groupG
. Prove the following:The order of
a^k
is a divisor (factor) of the order ofa
.
Approach 1
Let ord(a) = n
.
Let ord(a^k) = m
.
We must show that m
is a divisor of n
. I.e. m|n
.
Since ord(a^k) = m
we know that:
(a^k)^m = e
a^(km) = e
By theorem 5:
Suppose an element
a
in a group has ordern
.Then
a^t = e
ifft
is a multiple ofn
.("
t
is a multiple ofn
" means thatt = nq
for some integerq
).
km = nq
Solving for m
(the order of a^k
):
m = nq/k
It isn't clear that this shows that m
is a divisor of q
.
Any suggestions for this approach?
Approach 2
(This is the approach suggested by the hint in the back of the book.)
Let ord(a) = n
.
(a^k)^n
a^(nk)
(a^n)^k
e^k
e
At this point, the book says to use theorem 5.
By theorem 5:
nk = nq
Canceling n
on both sides:
k = q
And that's all the book's hint has to offer. What's a good way forward?
Note: there is a question on this site that specifically asks about Approach 2 shown above. What I'm asking here is, if Approach 1 is workable. It would also be nice to see Approach 2 completed as the full answer in the linked question was not shown.
Approach based on 2 above and Bill's answer
Let ord(a) = n
.
Let ord(a^k) = m
.
(a^k)^n
a^(nk)
(a^n)^k
e^k
e
Thus
(a^k)^n = e
Let's state theorem 5 in these terms
Suppose an element
a^k
in a group has orderm
.Then
(a^k)^n = e
iffn
is a multiple ofm
.