Let G be a group and let a be a fixed element of G. Definethe map a: G>G by a(g)

Question

Let G be a group and let a be a fixed element of G. Definethe map a: G>G by a(g)=ag forg G.

Explanation / Answer

Firstly, note that the map a is a bijective map onG as a set. Indeed, if a(g) = a(h), then by definition, ag=ah, so multiplying bya-1 on the left on both sides, we get g=h, soa is injective. Conversely, given any g in G,there exists a unique x in G, namely a-1g such thata(x) = g, so a is surjective aswell. Let SG denote the group of permutations of G (as aset). Now, we claim that the map G -> SG given by a ->a is an injective homomorphism. This will provewhat you want. First, is it a homomorphism? In other words, is ab= a b ? Here the group operationis composition of maps. ab (g) = abg= a(bg)=a(bg)=a (b(g)), so the map is indeed ahomomorphism. For injectivity, consider its kernel. Suppose a (g) = g for all g. We must show that a = 1 in G.a (g) = g => ag=g for all g in G, but clearlythen, a=1, so that the kernel of this group homomorphism is simply{1}. Hence it is injective.

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