By Randall R. Holmes

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1 Example We have Z ≤ Q ≤ R ≤ C, meaning each is a subgroup of the next (under addition). It is possible for one group to be a subset of another group without the former being a subgroup of the latter. For instance, R+ (positive reals) is a subset of R, and we have that R+ is a group under multiplication and R is a group under addition. But R+ is not a subgroup of R since R+ does not use the same binary operation as R. The following theorem provides the most common way to check whether a subset of a group is a subgroup.

A is a subgroup of G. Proof. 2). ) We have e = a0 ∈ a . ) Let x, y ∈ a . Then x = am and y = an for some m, n ∈ Z. So xy = am an = am+n ∈ a . ) Let x ∈ a . Then x = am for some m ∈ Z. So x−1 = (am )−1 = a−m ∈ a . By the Subgroup Theorem, a is a subgroup of G. We call a the (cyclic) subgroup of G generated by a. 2 Example (a) Find 3 in Q× (under multiplication). (b) Find 3 in Z (under addition). Solution (a) In Q× we have 3 = {3m | m ∈ Z} = {. . , 3−2 , 3−1 , 30 , 31 , 32 , . . } = {. . , 19 , 13 , 1, 3, 9, .

Ks ). Put I = {i1 , i2 , . . , ir } and K = {k1 , k2 , . . , ks }. To show that the two functions στ and τ σ are equal, we need to show that (στ )(m) = (τ σ)(m) for all m in their common domain, which is {1, 2, . . , n}. Let m ∈ {1, 2, . . , n}. First assume that m ∈ / I ∪ K (in other words, assume that m does not appear in either cycle). Then σ and τ both fix m giving (στ )(m) = σ(τ (m)) = σ(m) = m = τ (m) = τ (σ(m)) = (τ σ)(m). Now assume that m ∈ I. Then σ(m) ∈ I, as well. Since I and K are disjoint by assumption, m and σ(m) are not in K, so they are fixed by τ .