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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.

user26857
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Learnmore
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  • Note on the other hand that (like all groups) $A$ acts faithfully by permutation on itself, so $A \subset S_A \cong S_{#A} = S_6$. An old qualifying exam from my graduate program included the problem of determining this $n$ for all five groups of order $8$ up to isomorphism. It's quite a nice problem so you might like to attempt the other groups after tackling this case. IIRC the $Q_8$ case can be found as a question elsewhere here. – Travis Willse May 11 '15 at 01:23
  • $#A=8$, not $6$. – Greg Martin May 11 '15 at 01:23
  • Well, it's not hard to believe that $\langle (12)(34)(56) \rangle \cong A$, but showing that $S_6$ is minimal, if it even is, is the tricky part. – pjs36 May 11 '15 at 01:38
  • Well how to find this $n$ any hints @Travis – Learnmore May 11 '15 at 02:01
  • @GregMartin Oops, thanks for spotting the typo. I managed to conflate the intended hint and pjs36's. – Travis Willse May 11 '15 at 03:16
  • @learnmore Since we know that $4 \leq n \leq 6$, it's not too inefficient to proceed naively. For any such isomorphism $\phi$, $\phi(1, 0, 0)$ must have order $2$. In $S_4$ there are only two conjugacy classes whose elements have that order, with representatives $(12)$ and $(12)(34)$, so by conjugation we can assume that $\phi(1, 0, 0)$ is one of those. Now, $\phi(0, 1, 0)$ must be an element of order $2$ that commutes with $\phi(1, 0, 0)$. Which elements in $S_4$ commute with $(12)$? Which commute with $(12)(34)$? – Travis Willse May 11 '15 at 03:23

2 Answers2

8

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).

user26857
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  • how do you know the subgroups of $S_4/S_5$ ? – Learnmore May 11 '15 at 09:30
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    @learnmore As I said, it's about the subgroups with $8$ elements of $S_4/S5$. It's well known that $D_4$ is isomorphic to a subgroup of $S_4$ and thus it's a $2$-Sylow subgroup. It's also isomorphic to a subgroup of $S_5$ (since $S_4\subset S_5$), and a $2$-Sylow subgroup in $S_5$, too. But all $2$-Sylow subgroups are conjugated (hence isomorphic). – user26857 May 11 '15 at 09:41
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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 $H$.

C.S.
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