A (very) short introduction to Cyclic Groups
Juspreet Singh Sandhu
September 2018
In this short post, I plan on introducing Cyclic Groups, talking about the
Order of Cyclic Groups, and concluding with a theorem regarding the Sub-
Groups of Cyclic Groups. Note: In comparison to other posts, this one will be
shorter (and assume the basics of Groups). So, let us begin:
Def
n
: A group G =< S, · > is cyclic, iff, G =< a > , for some a ∈ G
Therefore, a Cyclic Group is one that can be generated by a single-element in
the Group.
Def
n
: A Cyclic Group C
m
has ord(C
m
) = ord(a) , where, C
m
=< a >
Intuitively, this means that the order of the Group is the number of elements
which is equivalent to the number of elements generated by the generator. It
is easy to show that all Cyclic-Groups with the same order are isomorphic to
each other.
Claim: Given C
i
, C
j
, , ord(C
i
) = ord(C
j
) , C
i
∼
=
C
j
Proof:
Let, C
i
= < a
c
i
> , C
j
= < a
c
j
>.
We can clearly construct a f : C
i
→ C
j
, , f((a
c
i
)
m
) = (a
c
j
)
m
,
m ∈ [1, ord(a
c
i
)]
It is easy to verify that this function is bijective, as the generator of one
group maps to the other and that means every element has precisely one
image.
Adjacency is preserved:
f((a
c
i
)
m
1
(a
c
i
)
m
2
)) = f((a
c
i
)
m
1
+m
2
) = (a
c
j
)
m
1
+m
2
= (a
c
j
)
m
1
(a
c
j
)
m
2
=
f((a
c
i
)
m
1
)f((a
c
j
)
m
2
)
This proves that any 2 groups of the same order are isomorphic.
QED
Now, we move to an extension of this idea: How does one deal with Cyclic
Groups of infinite order ?
Well, using the Integers.
Claim: Given C
i
, , ord(C
i
) = ∞, C
i
∼
=
Z
Proof:
The idea is similar to the first proof.
1
We can assign a bijection from < C
i
, · > → < Z, + > that shall simply do
the following 3 things:
i) f(e(C
i
)) = 0
ii) f(gen(C
i
)) = 1
iii) f(gen(C
−1
i
)) = −1
As one can easily see (using Induction), this will preserve adjacency, and it
is a bijective mapping.
This proves that any Cyclic Group of infinite order is isomorphic to the group
of Integers.
QED
Now that we’ve established the 2 most fundamental isomorphisms, we can now
look at Sub-Groups of Cyclic Groups. An intuitive use of the Remainder The-
orem will convince us that all Sub-Groups of a Cyclic Group are Cyclic.
Claim: Given C
i
, let S
‘
= {S | S <
s
C
i
}.
Then, S = < a
S
> , for some a
S
∈ C
i
, ∀ S ∈ S
‘
.
Proof:
Assume that some S ∈ S
‘
is not cyclic. This implies that ∃ some s
j
∈ S
‘
,
, s
j
= a
m
, where a is one of the generators of S. If this is the case, then
we see that, (s
k
j
) is also not generated by this element. However, all powers
of s
j
∈ S, due to closure of Groups. This leads to a contradiction, as
essentially, if this is not generated by the generator of C
i
, we will not be able
to retain the elements of powers of s
j
in S.
This can be seen formally by writing s
j
as:
s
j
= (a
C
i
)
bq+r
By definition, (a
C
i
)
bq
= e
So, s
j
= a
r
.
However, this is not possible because all elements of the cyclic sub-group
must be generated by the generator of the cyclic group.
QED
2