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Let $G$ be a finite group.

1.Suppose that $G$ acts transitively on a finite set $X$ with $|X|\geq2$.Show that there is an element $g\in G$ with no fixed points on $X$.

2.Let $H$ be a proper subgroup of $G$. Prove that

$G \neq$ $\cup_{x\in G} xHx^{-1}$

I have done the first part using Burnside Lemma. I am unable to do the second. The book asks to use the first part. I have no idea how to do that. Thanks!!!

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    Presumably the following is not the author's intention, but you can find another solution by simple counting argument in Dixon's Problems in Group Theory, [Problem 1.9](https://books.google.com/books?id=uoa1yO_Z_CIC&pg=PA5). – 2017-02-26
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    yeah I got the solution. Thanks for the help!! But I also want to do it the way the author asks to do it. You can see exercise 7.6 in the book "Representation of finite groups" by Benjamin Steinberg!! – 2017-02-26

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If $H$ is normal the result is trivial because $H$ exhausts the set of all conjugates, and $H$ is not the whole group. Otherwise note that $G$ acts transitively on the set of all conjugates $$\{xHx^{-1}|x\in G\}$$ The action is given by conjugation. An element $g$ with no fixed points for this action does not stabilize any of these conjugates, hence it cannot be contained in any of them. This is exactly what we wanted to prove (that such an element exists).

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    Note: for extreme (but trivial) case $H \trianglelefteq G$, the $G$-set has only one element and we cannot use 1. – 2017-02-26
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    @Orat Yes, I forgot that case. But that case is easy, I'll include it. – 2017-02-26
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    @MattSamuel Can you please specify the set on which the group is acting and the group action. Thanks! – 2017-02-26
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    @Riju I thought that was clear, but I tried to clarify a bit further in the edit. – 2017-02-26
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    @MattSamuel so you're telling that if g $\in xHx{-1}$ for some x$\in$G then g fixes $xHx^{-1}$ , since by the above problem there is atleast one such g which fixes nothing, so there will exist a g $\in$ G such that $g \notin xHx^{-1}$ for every x $\in$ G. Am i right? – 2017-02-26
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    @Riju Yes, since these are subgroups and closed under multiplication. – 2017-02-26