Tour:No proper nontrivial subgroup implies cyclic of prime order
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WHAT YOU NEED TO DO: Understand, thoroughly, the statement and proof of this theorem.
Statement
If a nontrivial group has no proper nontrivial subgroup, then it is a cyclic group of prime order. In other words, it is generated by a single element whose order is a prime number.
Converse
Further information: Prime order implies no proper nontrivial subgroup
The converse is clearly true from Lagrange's theorem. Namely, if we have a cyclic group of prime order , any subgroup must have order either 1 or
, by Lagrange's theorem. The only possible subgroup of order
is the whole group, and the only possible subgroup of order 1 is the trivial subgroup.
Note that this also shows something stronger: any group of prime order, since it has no proper nontrivial subgroup, must be cyclic of prime order.
- Cyclic of prime power order iff not generated by proper subgroups
- Every subgroup is a direct factor iff trivial or elementary Abelian
- Cyclic iff not a union of proper subgroups
Proof
Given: A nontrivial group , such that the only subgroups of
are the trivial subgroup, and
itself
To prove: There exists an element such that
, and the order of
is a prime number. In particular,
with
Proof: Since is nontrivial, there exists
in
. Then, consider the subgroup generated by
. This is a nontrivial subgroup, hence, by assumption,
.
Now there are two possibilities. First, that has infinite order: no positive power of
is trivial. In this case, the group
is isomorphic to (i.e., can be identified with) the group
, under the identification
. In particular, the subgroup generated by
, which corresponds to the even integers, is a proper nontrivial subgroup, leading to a contradiction.
Hence, cannot have infinite order, so let
be the order of
. Then,
. Suppose that
is composite. Then,
is isomorphic to the cyclic group of order
, under the identification
. Pick a positive integer
such that
. Then the subgroup generated by
is a proper nontrivial subgroup of
(corresponds to the multiples of
mod
). More precisely, it is a subgroup of order
, because the order of
is
. This is again a contradiction.
Hence, must be a prime, so
is cyclic of prime order (as desired).
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