62 Chapter 1. Groups I is just one group of order p. Of the possible orders through 7, four are primes, namely, 2, 3, 5, and 7, and so we need look only at orders 4 and 6. Proposition 1.97. Every group G of order 4 is isomorphic to either I4 or the four-group V. Moreover, I4 and V are not isomorphic. Proof. By Lagrange’s Theorem, every element in G has order 1, 2, or 4. If there is an element of order 4, then G is cyclic. Otherwise, x2 = 1 for all x ∈ G, so that Exercise 1.35 on page 27 shows that G is abelian. If distinct elements x and y in G are chosen, neither being 1, then we quickly check that xy / ∈ {1,x,y} hence, G = {1,x,y,xy}. It is easy to see that the bijection f : G → V, defined by f(1) = 1, f(x) = (1 2)(3 4), f(y) = (1 3)(2 4), and f(xy) = (1 4)(2 3), is an isomorphism, for the product of any two non-identity elements is the third one. We have already seen, in Example 1.60, that I4 ∼ = V. • Another proof of Proposition 1.97 uses Cayley’s Theorem: G is isomorphic to a subgroup of S4, and it is not too diﬃcult to show, using Table 1 on page 10, that every subgroup of S4 of order 4 is either cyclic or isomorphic to the four-group. Proposition 1.98. If G is a group of order 6, then G is isomorphic to either I6 or S3.30 Moreover, I6 and S3 are not isomorphic. Proof.31 By Lagrange’s Theorem, the only possible orders of nonidentity elements are 2, 3, and 6. Of course, G ∼ = I6 if G has an element of order 6. Now Exercise 1.36 on page 27 shows that G must contain an element of order 2, say, t. We distinguish two cases. Case 1. G is abelian. If there is a second element of order 2, say, a, then it is easy to see, using at = ta, that H = {1,a,t,at} is a subgroup of G. This contradicts Lagrange’s Theorem, because 4 is not a divisor of 6. It follows that G must contain an element b of order 3. But tb has order 6, by Proposition 1.89. Therefore, G is cyclic if it is abelian. Case 2. G is not abelian. If G has no elements of order 3, then x2 = 1 for all x ∈ G, and G is abelian, by Exercise 1.35 on page 27. Therefore, G contains an element s of order 3 as well as the element t of order 2. Now | s | = 3, so that [G : s ] = |G|/| s | = 6/3 = 2, and so s is a normal subgroup of G, by Proposition 1.68(ii). Since t = t−1, we have tst ∈ s hence, 30Cayley states this proposition in an article he wrote in 1854. However, in 1878, in the American Journal of Mathematics, he wrote, “The general problem is to find all groups of a given order n . . . if n = 6, there are three groups a group 1, α, α2,α3,α4,α5 (α6 = 1), and two more groups 1, β, β2,α,αβ,αβ2 (α2 = 1, β3 = 1), viz., in the first of these αβ = βα while in the other of them, we have αβ = β2α, αβ2 = βα.” Cayley’s list is I6, I2 × I3, and S3 of course, I2 × I3 ∼ = I6. Even Homer nods. 31 We give another proof in Proposition 4.87.

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