Showing a functor that takes a group to its set of subgroups is not representable Let $F:\text{Grp} \rightarrow \text{Set}$ be the functor that takes a group to its set of subgroups. Suppose $A$ is such a representing object, then $\text{Hom}(A,A) \simeq F(A) \Rightarrow \text{Hom}(S_{|A|},S_{|A|}) \simeq F(S_{|A|})$, i.e.
$Aut(S_{|A|}) \underset{\text{Set}}\simeq F(S_{|A|}),$ but $Aut(S_n) = S_n$ for all $n$ other than $2$ and $6$ and $|F(S_{|A|})|>|A|!$ for all $|A|>3$, and so we obtain a contradiction after eliminating the cases $2,3$ and $6.$
Is there (there has to be) a better, more structural proof of this?
 A: The first problem is that there are two possible such functors: one is covariant, and its effect on a morphism $G \to G'$ is to take a pair $H \subseteq G$ to $f(H) \subseteq G'$. However, there it could also take a pair $H'\subseteq G'$ to $f^{-1}(H') \subseteq G$. This one is the right one in most contexts (for instance in the category of topological spaces, where a pair $U\subseteq X$ could be an open subset). Let me call this functor $F$.
Suppose that there is a group $\Omega$ with the property that for any group $H$, there is a natural isomorphism
$$\hom(H, \Omega) \simeq F(H).$$
Then for every direct limit $\varinjlim H_i$, we have
$$F(\varinjlim H_i) = \hom(\varinjlim H_i, \Omega) \simeq \varprojlim \hom(H_i, \Omega) = \varprojlim F(H_i).$$
In other words a representable functor takes colimits to limits.
But take $$H_i = \frac{1}{p^i}\mathbf Z/\mathbf Z, \quad H_\infty = \varinjlim H_i = \frac{1}{p^\infty}\mathbf Z/\mathbf Z \subseteq \mathbf R/\mathbf Z.$$
The subgroups of $H_\infty$ are all of the form $\frac{1}{p^n}\mathbf Z/\mathbf Z$ for some $n\geq 0$, or the whole group. Thus $F(H_\infty)$ is countable. On the other hand the maps of the direct system $F(H_{i+1}) \to F(H_i)$, induced by inclusion of $\frac{1}{p^{i}}\mathbf Z/\mathbf Z$ in $\frac{1}{p^{i+1}}\mathbf Z/\mathbf Z$, identify in a natural way with
$$\dots \to \{0,1,2\} \to \{0,1\} \to \{0\}$$
where the map $\{0,1,2, \dots, i+1\} \to \{0,1, \dots, i\}$ takes $i+1$ to $0$ and maps every other element to itself. It is easy to see that the inverse limit of this system is uncountable. This shows that $F$ cannot be representable, because $F(H_\infty)$ and $\varprojlim F(H_i)$ do not have the same cardinality.
