I have a question on what it means for one group action to contain another group action as a normal subgroup. I am guessing this means that the image of the first group action contains the image of the second as a normal subgroup, but I want to double check.

The context where I saw this phrase is as follows. Given a group $G$, the right and left regular actions of $G$ on itself are defined by the maps $\rho: (x,g) \mapsto xg$ and $\lambda:(x,g) \mapsto g^{-1}x$, respectively. Then, the action of $G^* = G \times G$ on $G$, which is the product of the left and right regular actions, is defined by $\mu(x,(g,h))=g^{-1}xh$. The text I'm reading says

``This transitive action contains both the left and right regular actions as normal subgroups.''

My question is on what it means for one action to contain another as a normal subgroup. Since an action on $G$ corresponds to a homomorphism into $Sym(G)$, the image of this homomorphism (i.e. the permutation representation afforded by this action) could be what was meant. In fact, in this case, I verified that the image of the right regular action $\rho$ (and also of $\lambda$) is a normal subgroup of the image of the product action $\mu$ (all three of these images are subgroups of $Sym(G)$). So I am guessing what is meant by one action being a normal subgroup of another is that its image is a normal subgroup of the other image. Is that correct?

We can't always reduce or think of an action in terms of just its image since we lose some information if the action is not faithful. For example, two actions need not be isomorphic even if their images are. In fact, the definition of isomorphic actions is in terms of the action and not just in terms of the images of the actions.


Yes, I would presume that is the author's meaning. Essentially, this is just saying that $1 \times G$ and $G \times 1$ are normal subgroups of $G \times G$.

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  • $\begingroup$ In the question above, the two normal subgroups (the left and right regular representations $L(G)$ and $R(G)$, respectively) do not have a trivial intersection in general (their intersection has $|Z(G)|$ elements) and so the image of $\mu$ is not a direct product of $L(G)$ and $R(G)$, unlike in the case of $1 \times G, G \times 1 \trianglelefteq G \times G$. $\endgroup$ – AG. Aug 19 '13 at 8:41
  • $\begingroup$ That's true, but it doesn't matter. If $H$ is a normal subgroup of $G$, and $G$ acts on some set, then the image of $H$ under the action (even if not faithful) is a normal subgroup of the image of $G$. $\endgroup$ – Ted Aug 20 '13 at 2:01
  • $\begingroup$ I see. The product action $\mu$ of $G \times G$ on $G$, when restricted to the normal subgroup $G \times 1$, gives the left regular action; hence its image $L(G)$ is a normal subgroup of the image of $\mu$. $\endgroup$ – AG. Aug 21 '13 at 2:12

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