xp = x and xp = e. Note that SC is nonempty since it contains (e, e, ..., e). Th

Question

xp = x and xp = e. Note that SC is nonempty since it contains (e, e, ..., e). The result follows from the previous lemma since |SC| |S| 0 (mod p) and |SC| 0. If G is a p-group, then by Lagrange's theorem, the order of each element of G is a power of p. The standard definition of a p-group, allowing its usage for infinite groups, is that G is a p-group if each element of G has an order that is some power of p. Let G be a group acting on the set S, and let phi : G rightarrow Sym(S) be the group homomorphism defined in Proposition 7.3.2. Show that ker(phi) = x S Gx.

Explanation / Answer

To find the kernel: ifg?ker?,then?(g)=idS(where id_S is the identity function on S). Now, if?(g)=idSthen?(g)(x)=xfor allx?S.But, by definition of the action,?(g)(x)=gx. And thus, the condition is reallygx=x.But the "g" that satisfy that, are exactlyGx. Also, we want it to hold for allx?S, so we have to take the intersection?x?SGx.

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