# Verifying Group Homomorphism Preserves Action on a Set

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.

• 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 $\phi(e_G)x \in X$. – Lee Mosher Nov 24 '19 at 17:43

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.

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