Prove that .
We already know that by counting the possible images of, say, and . By a previous theorem, it suffices to show that no automorphism of has order 6.
Let be an automorphism of . Now has 6 elements of order 4, one element of order 2, and one identity; must fix the identity and the element of order 2, so that we can consider as an element of , where . Suppose now that some automorphism has order 6. Now , where , and , where . Clearly and . If , then and has order 3; thus , and we have . Now . Note that . If , then . Similarly, if , then . Both cases yield a contradiction, so that no automorphism of order 6 exists.
In addition, we record this handy lemma.
Lemma: Let be a group and let be characteristic. Then acts on by . Proof: First we need to show well-definedness: if , we have . Then since is characteristic in , so that . Moreover, and .