Group where you can see all the finite subgroups. Is there any group $G$ for which every finite group is a subgroup ?
I thing we can consider $S_\mathbb{N}$, because $S_n$ is subgroup of $S_\mathbb{N}$
for all $n$. so by Caylay's theorem we are done.
But I like to know, is there any other group which also having this property?
Thanks in Advance.
 A: The restricted permutation group $S_{(\mathbb N)}$ also works, of course; this is the set of permutations of $\mathbb N$ which fix almost all elements. This one is countable.
This group is also not minimal: the restricted permutation group of any infinite subset of $\mathbb N$ has, clearly, the same properties. There are therefore uncountably many subgroups with the same property, and even uncountable chains (order isomorphic to $\mathbb R$) of subgroups with the same property
A: The direct sum/product of (at least one copy but potentially more of) all finite groups. (Or of all directly indecomposable finite groups.) Given all the different ways to combine groups (direct products and sums, semidirect products, free products), there are an absolutely gargantuan number of ways to put the finite groups together into an infinite group (even if you fix its size).
One can use a "greedy algorithm" method of construction as well. First enumerate all finite groups in a listed sequence. Start with $G_0$ = the trivial group. At each step $n$, if the $n$th finite group is contained in $G_n$, then set $G_{n+1}=G_n$, otherwise form $G_{n+1}$ out of $G_n$ and that group (e.g. by picking a semidirect product to make). Then define $G$ to be the direct limit of these $G_n$s.
It might be an interesting question to characterize what might be "minimal" such groups.
A: So, the other answers are correct, but they are all lacking in one respect - size. What do I mean by "size"? Well, they lack finiteness properties. For example, $S_{\mathbb{N}}$ is uncountable, while the direct sum of all finite groups is infinitely presented.
"Monsterous" groups usually exist. If you dream up a reasonable* property you can normally find some group which satisfies it. Therefore, the trick is to find monsterous groups with additional finiteness properties. These properties are often actually not that strict - "finitely presented" is perhaps the most common. So, lets go for that one...
There actually exists a finitely presentable group which contains an isomorphic copy of every finite group. To see this, begin with anon's answer: enumerate all finite groups and take their direct sum. This gives you a group $Q_1$ with a recursive presentation, which by the Higman-Neumann-Neumann theorem embeds into a finitely generated, recursively presented group $Q_2$. One can then use Higman's embedding theorem to embed $Q_2$ into a finitely presented group $Q_3$, as required. (Please see Lyndon and Schupp's book Combinatorial group theory for the quoted theorems.)
(Note: why stick to finite groups? The same ideas prove that there exists a finitely presented group which contains an isomorphic copy of every finitely presented group, and in fact that there exists a finitely presented group which contains an isomorphic copy of every recursively presented group.)
*Even relatively unreasonable properties can be made to work. For example, for every prime $p>>1$ there exist infinite, finitely generated groups where every proper, non-trivial subgroup is cyclic of order this fixed prime $p$. It is an open problem as to whether there exist such groups which are finitely presented.
A: First, there is no single group that contains all finite groups as subgroups. This is so since there are simply too many finite groups, namely a proper class, and thus the class of all finite groups can't be contained in the power-set of any given set.
What is possible, as you and @anon show, is that there are groups that contain an isomorphic copy of each finite group. It is also possible to construct a group containing for every finite group a normal subgroup isomorphic to it. Clearly any such group must be infinite, it can't be abelian, and it must contain an element of any possible finite order. 
For any increasing sequence of natural numbers ${n_k}$, the group $\prod S_{n_k}$ contains an isomorphic copy of any finite group (though not necessarily as a normal subgroup). This group, just like the group $S_\mathbb N$, which has the same property, has cardinality $c$.
Call a group with the property we discuss a cool group. Now, it is tempting to consider the intersection of all cool groups, and hope to show that it is a cool group, and thus the smallest such. However, this is hopeless because the intersection will simply be the empty set. One can try to remedy this situation in various ways but I'm not sure if it leads to anything good. 
One obvious way to make the intersection make sense is to consider all cool groups that are subgroups of some fixed universal group. But then again, the up-to-isomorphism issue will prevent the intersection of all of these cool groups (which at least now is a group) from being cool. I'm not sure what else (if at all) there is to say about cool groups.
