Are all order 4 groups only isomorphic to C4 or C2+C2?

[Jesssa]

Is it correct to say that any order 4 group is only isomorphic to either
C4 or C2+C2 ?

where C4 is the order 4 cyclic group and C2 the order 2 cyclic group

 

[wisvuze]

Yes, there is C4 and the klein four group

 

[jgens]

There is actually a really cute proof of this result which can be extended to prove the following result: If p is a positive prime, then Z_{p^2} and Z_p \times Z_p are the only groups of order p^2 up to isomorphism. Anyway, on to the proof for groups of order 4 (I will leave the proof for groups of order p^2 to you).

Let G be a group of order 4 and let x \in G be an element of maximal order. Now consider the following two cases:

1. If |x| = 4, then G \cong Z_4.
2. If |x| = 2, then choose y \in G \setminus \langle x \rangle and notice that |y| = 2. A short argument shows that both G = \langle x \rangle \langle y \rangle and \langle x \rangle \cap \langle y \rangle = \{e\} hold; moreover, \langle x \rangle and \langle y \rangle are both normal subgroups of G since they both have index 2. This means that G \cong \langle x \rangle \times \langle y \rangle \cong Z_2 \times Z_2.

Since every element of maximal order in G must have either order 2 or order 4 by the Lagrange Theorem, this completes the classification of groups of order 4.

 

[lpetrich]

An alternate proof uses how multiplying a list of the elements of a group by an element creates a permutation of that list. The permutation either has all elements fixed, for e, or no elements fixed, for all the group's other element.

Let's consider G = {e,a,b,c}, where all the non-identity elements have order 2. If any of them have order 4, then the group is isomorphic to Z4.

For multiplying by a, e and a form one permutation cycle, and b and c a second permutation cycle. This is true for both left multiplication and right multiplication, something that means that the group is abelian. The group's overall multiplication table is thus
{{e,a,b,c}, {a,e,c,b}, {b,c,e,a}, {c,b,a,e}}

and it is equivalent to (aⁱ¹*b^j1) * (aⁱ²*b^j2) = (aⁱ¹⁺ⁱ²*b^j1+j2). Thus, the group is isomorphic to Z2 * Z2.