Determine if this is a left group action

Question

Let $G$ and $H$ be groups, and let $\phi: G \rightarrow H$ be a homomorphism of groups.

1. To each $g \in G$, associate a function $g: H \rightarrow H$ by $g(h)=\phi(g) h $. Determine if this is a left action of $G$ on $H$. Justify your answer.

2. To each $g \in G$, associate a function $g: H \rightarrow H$ by $g(h)=h\phi(g) $. Determine if this is a left action of $G$ on $H$. Justify your answer.

I guess I'm just confused what the questions are asking for. Are we supposed to find some function in terms of $g$ to create a left group action, or are we supposed to show $g$ is a left group action?

I know that by definition a group action on a set $H$ is a map $\alpha: G \times H \rightarrow H$ satisfying

i) $\alpha(gh,x)= \alpha (g, \alpha (h,x))$

ii) $\alpha(e,x)=x$.

Also, if $G$ acts on $H$, then the map $f: G \rightarrow S_H$ is defined by $f(g)=\tau_g$, where $\tau_g(x)=gx$ is a well defined homomorphism.

Thanks in advance!

Answer 1

The first one is a left group action, because

$$\begin{align} (g_1g_2)\cdot h&=\phi(g_1g_2)h\\ &=\phi(g_1)\phi(g_2)h\\ &=\phi(g_1)(g_2\cdot h)\\ &=g_1\cdot(g_2\cdot h) \end{align}$$ and $$e\cdot h=\phi(e)h=eh=h.$$

The second one isn't always a left group action, because

$$\begin{align} (g_1g_2)\cdot h &=h\phi(g_1g_2)\\ &=h\phi(g_1)\phi(g_2)\\ &=g_1\cdot h\phi(g_2)\\ &=g_2\cdot (g_1\cdot h), \end{align}$$ which in general is different from $g_1\cdot (g_2\cdot h)$, for instance when $\phi(g_1)$ and $\phi(g_2)$ don't commute.

Answer 2

The first subquestion is defining a map $\alpha: G\times H \rightarrow H$ by $\alpha : (g,h) \mapsto g(h)= \phi(g)h$ and is asking you to determine whether $\alpha$ satisfies the axioms of a left group action precisely as you've stated them.

The second subquestion defines a map that is slightly different. It defines a map $\beta : H \times G \rightarrow G$ by $\beta: (h,g) \mapsto g(h) = h\phi (g)$ and is similarly asking you to determine whether $\beta$ satisfies the axioms of a left group action.

I hope this clarifies the question a bit for you.