Find smallest positive integer $n$ such that this particular group is isomorphic to a subgroup of $S_{n} $ This is a problem of a masters exam for which I am preparing.

Find the smallest positive integer $n$ such that $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/2\mathbb{Z}$ is isomorphic to a subgroup of $S_{n}$.

The structure of $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/2\mathbb{Z}$ is easy to see.
But the problem arises on how can I be sure that a particular  group of permutations like $S_{4}$ or $S_{5} $ has a subgroup of element 8 or not.
Then only I can think of its structure.

So, my question is how I can be sure whether $S_{n} $ has a subgroup of specific order or not?

 A: $S_n$ needs to have three distinct commuting elements of order $2$, the product of no two of which is equal to the other. What are the elements of order $2$? In their disjoint cycle decomposition there are only transpositions. If two of the elements have transpositions that contain one index the same but not the other, then they do not commute. Thus we need all the transpositions in the cycle decomposition to either be identical or disjoint. For any two elements, we also need the symmetric difference of the two sets of transpositions (which is the product of the two elements) not to be equal to the third. Thus, we need each element to have a transposition in the cycle decomposition that the other two do not have. Thus we need at least three disjoint transpositions in the group, hence at least $6$ indices.
This gives a necessary condition; we need at least $S_6$. I'm sure you can find a subgroup of $S_6$ using this description that is isomorphic to the required group.
It's not enough for it to have a subgroup with $8$ elements. $S_4$ has a subgroup with eight elements. But it is not isomorphic to the group you seek.
A: It is very easy to see that $4\leq n\leq6$. Indeed we have $n\geq4$ because
$$|(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/2\Bbb{Z})|=8>|S_3|,$$
and we have $n\leq6$ because $S_6$ contains the subgroup generated by $(1\ 2)$, $(3\ 4)$ and $(5\ 6)$, which is isomorphic to $(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/2\Bbb{Z})$.

Now for some more structured observations assuming only the most elementary knowledge of group theory:
The first thing to note is that the order of every nontrivial element of $(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/2\Bbb{Z})$ is  $2$. Expressing elements of $S_n$ as products of disjoint cycles, we see that an element has order at most $2$ if and only if it is a nontrivial product of disjoint transpositions.
The next thing to note is that
$(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/2\Bbb{Z})$ is abelian.  So it remains to determine whether we can find $7$ products of disjoint transpositions in $S_n$ that commute with eachother.
We can simply list the elements. In $S_4$ these are
$$(1\ 2),\quad (1\ 3),\quad (1\ 4),\quad (2\ 3),\quad (2\ 4),\quad (3\ 4),\quad (1\ 2)(3\ 4),\quad (1\ 3)(2\ 4),\quad (1\ 4)(2\ 3).$$
A quick check shows that each of these commutes with only $2$ others. This shows that $n>4$.
We can do the same in $S_5$; a quick check shows that only
$$(1\ 2),\quad (3\ 4),\quad (3\ 5),\quad (4\ 5),\quad (1\ 2)(3\ 4),\quad (1\ 2)(3\ 5),\quad (1\ 2)(4\ 5),$$
commute with $(1\ 2)$, but not all $7$ elements commute with eachother. By symmetry this shows that such a subgroup cannot contain any transpositions. A further check among the products of $2$ disjoint transpositions shows that only
$$(1\ 2)(3\ 4),\quad (1\ 2)(3\ 5),\quad (1\ 2)(4\ 5),\quad (1\ 5)(3\ 4),\quad (2\ 5)(3\ 4),$$
commute with $(1\ 2)(3\ 4)$, so by symmetry such a subgroup doesn't contain any products of disjoint transpositions either. Then no such subgroup exists, so $n>5$.
A: $n\le 6$.
For n=4,5 We know $D_4$ is a non abelian subgroup of $S_n$ order 8 and it is 2-syllow. Any two syllow subgroup are mutually conjugate and conjugacy presearves commutavity but $D_4$ is not abelian.
(Two subgroup $K,H\in G$ are conjugate to each other means there is $g\in G$ such that $g^{-1}Hg=K$.)
Hence n=6
