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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?

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

2 Answers 2

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This is actually Cayley's theorem and it is true if $n$ is the order of $G$. The counter-example you suggest has $n = 8 < 24 = |G|$.

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The statement is true. You have references in the link you posted.

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    $\begingroup$ You have not given a counterexample. Your $f$ has yet to be defined. $\endgroup$
    – Plop
    Apr 18 at 10:43

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