Let \(G\) be a group and \(A\) be an abelian normal subgroup of \(G\). Show that \(\bar{G}\) acts on conjugation on \(A\) by \(\bar{g} \cdot a = gag^{-1}\), where \(g\) is any representative of the coset \(\bar{g}\). Give an explicit example to show that this action is not well defined if \(A\) is non-abelian.
The proof of the first part is rather straightforward and hence omitted.
Consider the normal subgroup \(A_5\) in \(S_5\); clearly \(A_5\) is not abelian since \((1\ 2\ 3\ 4\ 5)(1\ 2\ 4\ 3\ 5) = (1\ 3)(2\ 5)\) while \((1\ 2\ 4\ 3\ 5)(1\ 2\ 3\ 4\ 5) = (1\ 4)(2\ 5)\). Let \(g = (1\ 2)(3\ 4)\), \(h = (1\ 2)(3\ 5)\) and \(a = (1\ 2)(3\ 4)\). All of them are in \(A_5\), i.e. \(\bar{g} = \bar{h}\). However we have \(\bar{g} \cdot a = (1\ 2)(3\ 4)\) and \(\bar{h} \cdot a = (1\ 2)(4\ 5)\) which are not equal, so the action is not well-defined.
No comments:
Post a Comment