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.