Let us consider the group $A=\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_2$. Find the smallest positive integer $n$ such that $A$ is isomorphic to a subgroup of $S_n$.
My thought.
Since $o(A)=8$ then $n\geq 4$.
If $n=4$, then $8$ will divide $24$, but how to make sure whether it has an abelian subgroup of order $8$ or not since $A$ is abelian.
Any help.
On
Note :
$n\geq 4$ as you observed correctly.
If $H \cong \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$, then all elements in $H$ has to have order $2$. So all the elements in $H$ are either $2$ cycles or product of $2$ Cycles. Now, note that if $H$ contains a $2$ -cycle, then it has to be disjoint. (Why?) Think of an example. What if $H$ contains $(1\ 2)$ and $(2\ 3)$?
So we infer that our subgroup $H$ contains only disjoint $2$ cycles and product of disjoint $2$ cycles.
We now rule out the case $n=4$. Suppose $H$ contains the two cycle $(1 \ 2)$ Then $H = \{e, (1 \ 2), (3,\ 4), (1\ 2)( 3\ 4)\}$. Try out oter possibilities and see that $|H|=4$ always so $H \not\cong \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$.
By the same reasoning you can rule out $n=5$ as well.
For $n=6$, we can construct $H$ in the following way. $H=\{ e, (1\ 2), (3\ 4), (5,\ 6), (1\ 2)(3\ 4), (1\ 2)(5\ 6), (3\ 4)(5\ 6), (1\ 2)(3\ 4)(5\ 6)\}$
Now to prove $H$ is isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$ note that any abelian group of order $8$ has to be isomorphic to $\mathbb{Z}_{8}$ or $\mathbb{Z}_{4} \times \mathbb{Z}_{2}$ or $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$. The first two can't happen since there is no element of order $8$ and $4$ in $H$.
The smallest $n$ is $6$:
1. $A$ is isomorphic to $\langle(1,2),(3,4),(5,6)\rangle$.
2. For $n=4,5$ the only subgroup of order $8$ which $S_n$ does contain is the dihedral group $D_4$ (and its conjugates, being a $2$-Sylow subgroup).