Cayley's Theorem question: examples of groups which aren't symmetric groups. Basically, Cayley's Theorem says that every finite group, say $G$, is isomorphic to a subgroup of the group $S_G$ of all permutations of $G$.
My question: why is there the word "subgroup of"? If we omit this word, is the statement wrong? brief examples would be nice.
Thank you guys so much!
 A: For example, $D_4 \cong G$ where $G \leq S_4, G\not\cong S_4$.
The group $D_4$ has order $8$, and hence, there is no $S_n$ such that $|S_n| = n! = 8$. But there is a subgroup of $S_4$ which is isomorphic to $D_4$.
Furthermore, there exists lots of cyclic groups, but for $n > 2$, $S_n$ is NOT cyclic. But there are subgroups of $S_n$ that are cyclic. The same is true with respect to abelian groups, as DonAntonio points out. 
So the theorem would be absurd if the restriction "isomorphic to a subgroup of $S_G$" were omitted!
A: If you take any group of order $n>2$, the number of elements of $S_G$ is $n!$, which is strictly greater than $n$, so there is no way that $G$ can be isomorphic to $S_G$. What this means is that there are permutations of the elements of $G$ that can't be realised simply by multiplying by a group element.
A: The symmetric group $S_n$ has order $n!$ whereas there exists a group of any order (eg. $\mathbb{Z}_n$ has order $n$).
A: Hint $ $ In $\rm\,\langle \Bbb Z/n, +\rangle\,$ if cycle $(0,1)$ is a shift $\rm\:x\to x\!+\!a\:$ then $\rm\: 0\!+\!a\equiv1,\,\ 1\!+\!a\equiv 0\:\Rightarrow\:1\equiv a\equiv -1.$
A: Consider the group $\langle \Bbb Z, +\rangle$ of integers under addition.
The Cayley representation of this group contains permutations $P_a$ where $$P_a(n) = n+a.$$  That is, each element of the Cayley representation is a permutation of $\Bbb Z$ obtained by shifting the entire number line rigidly left or right . (Or, in the case of $P_0$, leaving it fixed.)
Clearly, there are many permutations of $\Bbb Z$ that are not of this form.  For example, one such permutation exchanges $17$ and $23$ and leaves all the other integers fixed; this is not an element of the Cayley representation of 
$\langle \Bbb Z, +\rangle$.
