By Randall R. Holmes

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Hence, ϕ satisfies the homomorphism property. We have shown that ϕ is an isomorphism, so we conclude that G ∼ = Zn . 5 Direct sum of cyclic groups The following example illustrates one implication in the main theorem of this section. 1 Example Show that Z2 ⊕ Z3 ∼ = Z6 . Solution Put x = (1, 1) ∈ Z2 ⊕ Z3 . We have 1x = (1, 1) 2x = x + x = (0, 2) 3x = 2x + x = (1, 0) 4x = 3x + x = (0, 1) 5x = 4x + x = (1, 2) 6x = 5x + x = (0, 0) = e, 56 so x = Z2 ⊕ Z3 . This shows that Z2 ⊕ Z3 is cyclic. 1(ii), we get Z2 ⊕ Z3 ∼ = Z6 , as claimed.

Any element of Sn can be written as a product of transpositions. Note: The case n = 1 is included by allowing the possibility of a product with no factors, which is interpreted to be the identity. ) Proof. 2 it suffices to show that every cycle can be written as a product of transpositions. Let σ = (i1 , i2 , . . , ir ) be a cycle. We proceed by induction on r. If r = 1, then σ is the identity, which, according to our convention, is a product of transpositions with no factors. Assume that r > 1.

Since (xy)z = x(yz) for every x, y, z ∈ G, this equation holds for every x, y, z ∈ H, so (G1) is satisfied. The identity e of G is in H by property (i) and this same element acts as an identity element of H, so (G2) is satisfied. Finally, if x ∈ H, then its inverse x−1 (which exists in G) is actually in H by property (iii) and since x−1 x = e and xx−1 = e, (G3) is satisfied. Therefore, H is a group. The group G is a subgroup of itself. If H is a subgroup of G and H = G, then H is a proper subgroup of G.

### Abstract Algebra I by Randall R. Holmes

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