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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 $L_e=G$

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  • $\begingroup$ 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. $\endgroup$
    – anomaly
    Apr 6, 2017 at 21:21
  • $\begingroup$ @anomaly Now this makes sense. Thanks $\endgroup$
    – user330477
    Apr 6, 2017 at 21:22
  • $\begingroup$ 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. $\endgroup$
    – freakish
    Apr 6, 2017 at 21:33
  • $\begingroup$ @freakish Can you please give me a hint $\endgroup$
    – user330477
    Apr 6, 2017 at 21:42
  • $\begingroup$ 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 $g$. $\endgroup$
    – anon
    Apr 6, 2017 at 21:59

2 Answers 2

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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.

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  • $\begingroup$ Can you please elaborate $\endgroup$
    – user330477
    Apr 6, 2017 at 22:02
  • $\begingroup$ @user330477 On which part? What don't you understand? $\endgroup$
    – anon
    Apr 6, 2017 at 22:02
  • $\begingroup$ The function and perm(G) part $\endgroup$
    – user330477
    Apr 6, 2017 at 22:03
  • $\begingroup$ 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? $\endgroup$
    – anon
    Apr 6, 2017 at 22:04
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    $\begingroup$ 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? $\endgroup$
    – anon
    Apr 6, 2017 at 23:44
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Hint: you need to prove that

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

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

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