Sentence from Dummit and Foote

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Homework Statement



"Two group elements induce the same permutation on ##A## if and only if they are in the same coset of the kernel (if and only if they are in the same fiber of the permutation representation). In particular an action of ##G## on ##A## may also be viewed as a faithful action of ##G/\ker \varphi## on ##A##."

Homework Equations

The Attempt at a Solution



I am having trouble parsing the quotation given above, which comes from Dummit and Foote. Letting ##\varphi : G \to S_A## defined by ##\varphi(g) = \sigma_g## denote the permutation representation, does the first sentence in quotation say "If ##g,h \in G##, then ##\sigma_g = \sigma_h## if and only if ##g,h \in x \ker \varphi## for some ##x \in G## if and only if ##g,h \in \varphi^{-1}(\sigma_y)## for some ##\sigma_y \in S_A##"?

As for the second sentence, it seems to say that the group action of ##G## on ##A## induces a faithful action of ##G/\ker \varphi## on ##A##. What exactly is this induced action? The best I could come up with is ##(g \ker \varphi ) \cdot a = g \cdot a##. Is this right?
 
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Yes and yes. The elements of the kernel are those who map on the identity permutation of ##A## which means they don't contribute to the action. Simultaneously this destroys uniqueness, resp. a faithful action. Dividing it off solves the problem, which of course has to be shown, because such a quotient doesn't always work, e.g. ##C(G/C(G)) \neq_{i.g.} \{1\}##, i.e. the quotient of a group and its center doesn't necessarily lead to a group with a trivial center.