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Order 4 group isomorphic to

  1. May 23, 2012 #1
    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
  2. jcsd
  3. May 23, 2012 #2
    Yes, there is C4 and the klein four group
  4. May 23, 2012 #3


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

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

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

    Since every element of maximal order in [itex]G[/itex] must have either order [itex]2[/itex] or order [itex]4[/itex] by the Lagrange Theorem, this completes the classification of groups of order [itex]4[/itex].
  5. May 23, 2012 #4
    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*bj1) * (ai2*bj2) = (ai1+i2*bj1+j2). Thus, the group is isomorphic to Z2 * Z2.
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