I am struggling with this.
We know that every finite group $G$ is isomorphic to a subgroup of some symmetric group $S_n$ (Cayley's theorem).
But in the symmetric group $S_n$, every element - permutation can be written as a product of transpositions (cycles of length 2).
Order of a transposition is $2$. (For transposition $\pi$ this means $\pi ^2 = id$ = identity.)
So every element of the group $G$ can be written as a product of elements $a \in G$ with order $2$ ($a^2 = 1$).
But this obviously can not hold as for example a cyclic group with an odd number of elements has no other element but $1$ for which it would be $a^2 = 1$.
Where is the mistake?
Isomorphism preserves the order of elements because it is injective.
EDIT:
I imagine something like this:
$G = \mathbb{Z_4} = \{0, 1, 2, 3\}$
We have a homomorphism $\Phi: \mathbb{Z_4} \rightarrow S_4$
(We get it using Cayley's theorem)
$\Phi (0) = id$
$\Phi (1) = (0123)$ that can be rewritten as $(03)(02)(12)$
$\Phi (2) = (02)(13)$
$\Phi (3) = (0321)$ that can be rewritten as $(13)(23)(01)$
$\Phi$ is isomorphism on its image as it is already injective.
Now we have an isomorphism $\Psi$ between $\mathbb{Z_4}$ and $\{id, (03)(02)(12), (02)(13), (13)(23)(01)\}$, which is a subgroup of $S_4$.
$\Psi ((02)(13)) = \Psi ((02)) \cdot \Psi ((13))$
$\Psi ((02)) $ and $ \Psi ((13)) $ have the order $2$. So we have written $2 = \Psi ((02)(13))$ as the product of two elements with order $2$.
F or $1 = \Psi ((03)(02)(12)) = \Psi ((03)) \Psi ((02)) \Psi ((12))$ Now it looks like we have written $1 \in \mathbb{Z_4}$ as the product of three elements of order $2$.
Every finite group is isomorphic to a subgroup of some $S_n$,
but the transpositions in $S_n$ are not necessarily in that subgroup.