Proof for a subgroup $H$ of the finite subgroup $G$ there are at least $|H|$ elements that are not in conjugates of $H$ A problem in Isaac's finite group theory is to prove if $G$ is a finite group and $H$ is a proper subgroup of $G$ then there are at least $|H|$ elements that are not of the form $ghg^{-1}$ with $g\in G$ and $h\in $H.
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 A: Let $\chi$ be the permutation character associated with the action of $G$ on the right cosets of $H$ (by right multiplication). This action is clearly transitive, hence by 1A.6 in the same book
$$\sum_{g \in G}\chi(g)=|G|.$$
Let $g \in G-H$. Under the induced action of $H$ on the right cosets of $H$, $H$ and $Hg$ are in different orbits. So there are at least $2$ orbits and we conclude that $$\sum_{h \in H}\chi(h) \geq 2|H|.$$
again by 1A.6.
Let $S=\{g \in G : \chi(g)=0\}$. Since $\chi(g)=\#\{Hx:Hx=Hxg\}=\#\{Hx:xgx^{-1} \in H\}=\#\{Hx:g \in H^x\}$, $S$ is exactly the set of elements in $G$ that are not contained in any conjugate of $H$.
Observe that for every $h \in H$, $\chi(h) \neq 0$. This implies, $H \subseteq G-S$. Now,
$$\sum_{g \in G}\chi(g)=\sum_{g \in G-S}\chi(g)=\sum_{g \in H}\chi(g)+\sum_{g \in (G-S)-H}\chi(g) \geq 2|H|+|(G-S)-H)|.$$
Since $|(G-S)-H)|=|G -(S \cup H)|=|G|-|S \cup H|$, we get $|S \cup H|\geq 2|H|$.  Now finally note that $|S| + |H|\geq|S \cup H|$, yielding $|S| \geq |H|$, which is what you wanted to prove.
Additional Remark If we apply this to $H$ being a Sylow $p$-subgroup $P$, then we obtain $$|G -\bigcup_{g \in G}P^g| \geq |P|.$$ But more is true in fact. It is easy to see that 
$$\bigcup_{g \in G}P^g=\{x \in G : x \text { has $p$-power order}\}=\{x \in G : x^{|P|}=1\}.$$ A theorem of Frobenius, dating from 1907  (see for example http://bit.ly/101FCCu), implies that the cardinality of this last set $\{x \in G : x^{|P|}=1\}$ is divisible by $|P|$. Since $G$ is a disjoint union and $|P| \mid |G|$:
$$G = \bigcup_{g \in G}P^g \cup (G -\bigcup_{g \in G}P^g)$$
we obtain $|G -\bigcup_{g \in G}P^g| \equiv 0 \text { mod } |P|$.
