# Yes/No: There always exists an injective homomorphism from $G$ into $S_n$. [closed]

Let $$G$$ be a finite group of order $$n\ge2$$. Is the following statements true/false?

There always exists an injective homomorphism from $$G$$ into $$S_n$$.

My attempt: I found the answer here. I think this statement is False.

Take $$G=\mathbb{Z}_{24}$$ and $$S_8$$.

Now $$f:\mathbb{Z}_{24} \to S_8$$ is not injective.

Here I take $$n=8$$ and $$f$$ denotes group homomorphism.

Edit: $$f:\mathbb{Z}_{8} \to S_8$$ is not injective.

Am I right?

• What is $f$?$\,$
– Gary
Apr 18 at 10:37
• consider the $G$ action on $G$ by left multiplication, which gives a homomorphism $G\to S_n$.
– SRA
Apr 18 at 10:41
• But what map is $f$ exactly?
– Gary
Apr 18 at 10:53
• But there are several homomorphism. Which one is your $f$? Why is it not injective? Where did you prove that it isn't?
– Gary
Apr 18 at 10:55
• See this post for a counterexample for smaller $n$ than the order of $G$. Apr 18 at 11:06

This is actually Cayley's theorem and it is true if $$n$$ is the order of $$G$$. The counter-example you suggest has $$n = 8 < 24 = |G|$$.
• You have not given a counterexample. Your $f$ has yet to be defined.