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Let $G_1, G_2$ be two groups that act on a set $S$ on the left, such that for all $g \in G_1$ there's $g' \in G_2$ such that $g\cdot s = g'\cdot s$ for all $s \in S$.

Define $h : G_1 \to G_2$, $h(g) = g'$ as defined above. Then $h(ab) = (ab)'$. Well $(ab)\cdot s = (ab)'\cdot s \ \forall s \implies \text{ it equals } a(b\cdot s) = a'(b'\cdot s)$. But I can't conclude that $(ab)' = a'b'$.

How are the two groups related?

Thanks @ThomasAndrews. "If the group actions are faithful then $G_1 \approx G_2$."

Let $h$ be well-defined by choosing any such $g'$ for each $g \in G_1$. From the above, as $(ab)' \cdot s = a'(b' \cdot s)$, we have that $b'^{-1} a'^{-1} (ab)' = 1$, so that equality follows. That means if $G_2 \times S \to S$ is faithful then $h$ is a homom. $h$ is injective as $g'\cdot s = h' \cdot s \ \forall s$, then by faithfullness of $G_2$ again $g' = h'$.

Intuitively, if we have the similar requirement and a map $h' : G_2 \to G_1$, then $h$ becomes an isomorphism.

Daniel Donnelly
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    That $h$ function is not well-defined, because there can be multiple such $g'$, at least as far as your definition works. – Thomas Andrews Apr 20 '14 at 01:44
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    If the group actions are "faithful," then $G_1$ and $G_2$ are isomorphic. http://mathworld.wolfram.com/FaithfulGroupAction.html – Thomas Andrews Apr 20 '14 at 01:46

1 Answers1

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If the group actions are faithful, then Thomas Andrews's comment answers the question. For general group actions, the situation is as follows. In each of your two groups $G_i$, the subset $H_i$ of elements that act as the identity on $S$ (i.e., map every element of $S$ to itself) is a normal subgroup. The two quotient groups $G_i/H_i$ are isomorphic.

Andreas Blass
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