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I'm going through "A First Course in Abstract Algebra," by Fraleigh, and I just started going through the chapters on actions on sets. I came across an exercise that asked about two groups related by a homomorphism and the implications of actions on a set:

Let $\phi:G\rightarrow H$ be a group homomorphism, and let $H$ on act a set $X$. What does this tell us about whether or not $G$ acts on $X$?

I think that it does imply that $G$ acts on $X$ (it seems rather intuitive), and I've attempted to prove it below:

If a set $H$ acts on a set $X$, then we have that:

$(i)$ $\forall x\in X$, $e_Hx=x$

$(ii)$ $(h_1h_2)(x)=h_1(h_2x)$, $\forall x\in X$, $h_1,h_2\in H$

To show $(i)$, we note that $e_H=\phi(e_G)$, which implies that:

$$\phi(e_G)x=x$$

for all $x\in X$.

To show $(ii)$, we note that by the fact that $\phi$ is a homomorphism:

$$\phi(g_1g_2)=\phi(g_1)\phi(g_2)$$

for $g_1,g_2\in G$. We also have that $\phi(g_1)=h_1$, and $\phi(g_2)=h_2$. So:

$$\phi(g_1)\phi(g_2)=\phi(g_1g_2)=(h_1h_2)\rightarrow\phi(g_1g_2)(x)=h_1(h_2x)$$

by substitution into our previous statement. But the RHS is just:

$$h_1(h_2x)=\phi(g_1)\big[\phi(g_2)x\big]$$

so:

$$\rightarrow\phi(g_1g_2)(x)=\phi(g_1)\big[\phi(g_2)x\big]$$

Does this suffice/is there anything that can be made clearer/more concise? It seems like a fairly straightforward exercise, but the way it was phrased makes me a tad skeptical that it truly is that simple. Any advice would be sincerely appreciated. Cheers.

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  • $\begingroup$ By the way, your equations $e_Hx=x$ and $\phi(e_G)x=x$ look suspicious. It seems that your intention was to write down the definition of group actions, in which case you should have written $e_H x \in X$ and $\phi(e_G)x \in X$. $\endgroup$ – Lee Mosher Nov 24 '19 at 17:43
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Your intuition is spot on, and the proof is okay (although I would write all the details to be sure of what's going on).

To be explicit, the action $G \circlearrowright X$ is defined as

$$ g \cdot x := \phi(x) \cdot x, \tag{1} $$

where the second dot means the product in the $H$-action.

As for shorter proofs, if you have seen that actions $H \circlearrowright X$ are in correspondence with group homomorphisms $H \xrightarrow{\alpha} S(X)$ from $H$ to the group of bijections of $X$, then this reasoning reduces to noting that given a morphism $\phi : G \to H$ the composition

$$ G \xrightarrow{\phi} H \xrightarrow{\alpha} S(X) $$

is a group morphism, defining thus an action on $X$.

If you unwind the proof of this correspondence, you will obtain $(1)$ once again.

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  • $\begingroup$ Great, thank you for your response! $\endgroup$ – scoopfaze Nov 24 '19 at 6:39

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