By Antonio Machì

Introductory Notions -- basic Subgroups, Conjugation and Isomorphism Theorems -- workforce activities and Permutation teams -- turbines and kinfolk -- Nilpotent teams and Solvable teams -- Representations -- Extensions and Cohomology -- technique to the workouts

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**Example text**

Conversely, if H ≤ G and |H| = m, let H = ak , k ≤ n. Since o(ak ) = m, and o(ak ) = n/(n, k), we have m = n/(n, k), (n, k) = n/m, so that n/m divides k. Let k = (n/m)t; then n n n ak = a m t = (a m )t ∈ a m , n and therefore ak ⊆ a m . But both these subgroups have order m, and one being contained in the other they coincide. ♦ This result admits the following converse: if a ﬁnite group admits at most one subgroup for every divisor of the order, then it is cyclic (and therefore, by the theorem, it has has only one such subgroup).

Hence, two elements of G have the same image under ϕ if, and only if, they belong to the same coset of the kernel of ϕ. In other words, the equivalence relation “aρb if ϕ(a) = ϕ(b)” coincides with the relation “aρb if ab−1 ∈ ker(ϕ)”, and its classes are the cosets of ker(ϕ). The mapping ϕ(a) → Ka between the elements of the image of ϕ and the elements of the quotient group G/K, K = ker(ϕ), is therefore one-to-one. Moreover, this mapping is a homomorphism, because if ϕ(a) → Ka, ϕ(b) → Kb then ϕ(a)ϕ(b) = ϕ(ab) → Kab = KaKb.

E. hk = kh. This result may also be proved by considering the element h−1 k −1 hk. If we read it as h−1 (k−1 hk), then it is a product of two elements of H because, by the normality of H, k −1 hk ∈ H; if we read it as (h−1 k −1 h)k) it is the product of two elements of K, by normality of K. Hence h−1 k −1 hk = 1, and hk = kh3 . ♦ In the quaternion group, the subgroups H = i and K = j are normal, and therefore commute, but not elementwise. Indeed, ij = k and ji = −k. We have |H ∩ K| = 2, and therefore the assumption |H ∩ K| = 1 cannot be removed.