By Mark Steinberger

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Once again, there is a deviation from commutativity between the Sθ and a. The following lemma may be veriﬁed by direct computation. 3. We have Sθ · a = a · S−θ . Because Sθ and Sφ commute for any real numbers θ and φ, we obtain the following corollary. 4. The matrix a is not equal to Sθ for any θ. Write b (or b2n if more than one quaternionic group is under discussion) for S2π/2n . Then b has order 2n. 5. The quaternionic group of order 4n, Q4n ⊂ Gl4 (R), is given as follows: Q4n = {bk , abk | 0 ≤ k < 2n}, with a and b as above.

Let n be a positive integer. Then there are groups of order n. 4, Zn = Z/ ≡, where ≡ denotes equivalence modulo n. CHAPTER 2. 9. The canonical map π : Z → Zn is deﬁned by setting π(k) = k for all k ∈ Z. 4 for an arbitrary equivalence relation. We shall see in the next section that the canonical map π : Z → Zn is what’s known as a homomorphism of groups. 10. 1. Show that Zn is generated as a group by 1. Thus, Zn is a cyclic group. 2. Show that 2 does not generate Z4 . Deduce that there are ﬁnite cyclic groups that possess nontrivial proper subgroups.

Let f : G → G be a homomorphism. (a) Let H ⊂ G be a subgroup. Deﬁne f −1 (H ) ⊂ G by f −1 (H ) = {x ∈ G | f (x) ∈ H }. Show that f −1 (H ) is a subgroup of G which contains ker f . (b) Let H ⊂ G be a subgroup. Deﬁne f (H) ⊂ G by f (H) = {f (x) | x ∈ H}. Show that f (H) is a subgroup of G . If ker f ⊂ H, show that f −1 (f (H)) = H. (c) Suppose that H ⊂ im f . Show that f (f −1 (H )) = H . Deduce that there is a one-to-one correspondence between the subgroups of im f and those subgroups of G that contain ker f .

### Algebra [Lecture notes] by Mark Steinberger

by David

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