# Then prove that $g \rightarrow L_g$ is a group homomorphism.

For any group $G$ any $g \in G$, let $L_g:G \rightarrow G$ be the function $L_g(h)=gh$ and $R_g:G \rightarrow G$ be the function $R_g(h)=hg$. Let $S_G$ be the group of permutations of the group $G$ considered only as a set.

Then prove that $g \rightarrow L_g$ is a group homomorphism. What is the kernel of this homomorphism?

Doubt:

Can someone clarify what the question is trying to ask. The notation seems wrong to me. Now clarified

My attempt:

$L_g(h_1h_2)=g(h_1h_2)$ and $L_g(h_1) \cdot L_g(h_2)=(gh_1) \cdot (gh_2)$

How do I show $L_g(h_1h_2)=L_g(h_1) \cdot L_g(h_2)$

What is the kernel of this group homomorphism?

I think it is $G$ because $e$ is the identity of $G$ and $e$ gets mapped to $L_e=G$

• What exactly is unclear? You're trying to show that the map $G \to S_G$ given by $g \to L_g$ is a homomorphism. Apr 6, 2017 at 21:21
• @anomaly Now this makes sense. Thanks Apr 6, 2017 at 21:22
• Normally you would use different arrow for a definiton of a mapping: $g\mapsto L_g$ unlike $G\to S_G$ which only defines a domain and codomain. Apr 6, 2017 at 21:33
• @freakish Can you please give me a hint Apr 6, 2017 at 21:42
• You are not trying to show that $L_g$ is a group homomorphism for each $g$, you are trying to show that $g\mapsto L_g$ is a group homomorphism as a function of $g$.
– anon
Apr 6, 2017 at 21:59

For each $g\in G$ we have a left-multiplication function $L_g$ defined by $L_g(x)=gx$.

It is not true that $L_g$ itself will be a group homomorphism. In fact, since $L_g(e)=g$ and group homomorphisms preserve identities, it will only be a group homomorphism if $g=e$.

Consider the function $\mathrm{Left}:G\to \mathrm{Perm}(G)$ (where $\mathrm{Perm}(G)$ is the group of permutations of $G$'s underlying set) defined by $\mathrm{Left}(g)=L_g$. You want to show that $\mathrm{Left}$ is a group homomorphism.

• Can you please elaborate Apr 6, 2017 at 22:02
• @user330477 On which part? What don't you understand?
– anon
Apr 6, 2017 at 22:02
• The function and perm(G) part Apr 6, 2017 at 22:03
• Sorry, forgot to write what $\mathrm{Left}$ is. It is the function which takes $g$ as input and outputs $L_g$. As for the $\mathrm{Perm}(G)$ part, surely you know what a permutation is?
– anon
Apr 6, 2017 at 22:04
• The homomorphism we're talking about is from $G$ to $\mathrm{Perm}(G)$. The group operation in $\mathrm{Perm}(G)$ is function composition. The kernel will be the set of all $g\in G$ that get mapped to the identity element of $\mathrm{Perm}(G)$, in other words the identity function. What does it mean for $L_g$ to be the identity function? It means $L_g(x)=x$ for all $x$. In other words, $gx=x$ for all $x$. For what values of $g$ is that true?
– anon
Apr 6, 2017 at 23:44

Hint: you need to prove that

$$L_{g_{\large 1}g_{\large 2}} = L_{g_{\large 1}}\circ L_{g_{\large 2}}.$$

• Thanks. I think this clears everything up. Apr 6, 2017 at 22:06
• What is the kernel of homomorphism. I think it is just $G$ Apr 6, 2017 at 22:41
• @user330477, $g$ is is in the kernel iff $L_g =$ identity iff... Apr 7, 2017 at 6:11