The notation is just notation. If you like better
or
in lieu of
. They all mean the same thing except that
is reserved for when the groups are abelian (btw
is an analogous index operator for
as that of
for
).
Ok, so for the idea. You need a quick little lemma.
Lemma: Let
denote a cyclic (this lemma will show
the cyclic) group of order
. Then,
Proof: Let
be generated by
and define
by
.
Clearly
is injective since
=\theta(m)\implies c^k=c^m)
now we may assume WLOG that
and so the above implies that
. But!
and so
can't be strictly positive and since it's non-negative it follows that
Surjectivity is clear since
for some
and so
and
=c^k=g)
The homomorphism property follows since
=c^{m+k}=c^mc^k=\theta(m)\theta(k))
.
The conclusion follows.
Thus, we must only prove that
is cyclic and it will follow from the lemma that it is equal to
To see this we will prove that
)
generates
. To do this note that
|)
is the least positive number such that
^d=(d,\cdots,d)=(0,\cdots,0)\quad{\col or{red}(1)})
. But, notice that the first coordinate's order is
and so it will only be
when
where
and
is as in
)
. Similarly, the second slot will only be
when
and so on and so forth. Thus,
is the first occurrence of the multiples of
matching up. In other words
|=\text{lcm}(m_1,\cdots,m_k))
. But, since
=1)
and
=\frac{m_1\cdots m_k}{\text{gcd}(m_1,\cdots,m_k)})
it follows that
|=\text{lcm}(m_1,\cdots,m_k)=m_1\cdot s m_k)
. And since
it follows that
)
generates it. Thus,
is a cyclic group of order
and thus
by our lemma.
