$D_4$ subgroup of $S_4$

Question

I know similar questions have been asked, but this is not a duplicate!

I'm looking for a subgroup of $S_4$ isomorphic to $D_4$. This is not an assignment question, I am just studying for an upcoming test. Now, typical answers to this question go along these lines: take $r=(1234)$ and $f=(24)$. Then $r$ and $f$ are seen to satisfy the group relations of $D_4$, so they generate the desired subgroup. Here is my issue, so far we have defined a group morphism $D_4 \rightarrow S_4$, but how do we know this is injective? We haven't proven that $r$ and $f$ aren't subject to more relations, so it seems to me like this subgroup might be smaller than $D_4$. Intuitively, I know this isn't true, and of course I could by hand work out every element in the group generated by $r$ and $f$ and proceed from there, but I'm hoping for a more geodesic solution.

Answer 1

For those interested, here is what I came up with.

Let $r=(1234)$ and $f=(12)$. Clearly $r$ and $f$ satisfy the group relations of $D_4$, i.e. $r^4=e=f^2$ and $rf=fr^{-1}$. But how do we know that $r$ and $f$ are not subject to more relations? Let $H \leq S_4$ be the subgroup generated by $r$ and $f$. We certainly know that $|H| \leq 8$, and the proof will be complete when we show that $|H|=8$.

Clearly $f \neq r$. Now, $r^2=(43)(42)(41)(43)(42)(41)$ is an even permutation. $f$ is odd, so $f \neq r^2$. Next, note that $r^3=r^{-1}$, so $|r^{3}|=|r|=4$ while $|f|=2$. So, $f \neq r^3$. We conclude that $f \notin \{e, r, r^2, r^3\}$. Therefore $H$ contains an order 4 subgroup. So, $|H|$ is a multiple of 4. But $f$ does not belong to this order 4 subgroup, so $|H| > 4$. Recalling that $|H| \leq 8$, the only possibility is that $|H| = 8$, completing the proof.

Answer 2

You can concretely build it up from the scratch this way.

Let $D_4=\langle r,s\mid r^4=s^2,rs=sr^{-1}\rangle$. Now, $H:=\{1,s\}$ is a subgroup of $D_4$. The action of $D_4$ by left multiplication on the left quotient set $D_4/H$ is faithful, as its kernel is the intersection of all the cojugates of $H$ in $D_4$, which is certainly contained in the two terms intersection: \begin{alignat}{1} H\cap rHr^{-1} &= \{1,s\}\cap\{1,rsr^{-1}\} \\ &= \{1,s\}\cap\{1,r^2s\} \\ &= \{1\} \end{alignat} because $r^2\ne 1$. So we have an injective homomorphism: $$\varphi\colon D_4\to S_{D_4/H}, \space g\mapsto (g'H\mapsto gg'H)$$ In order to make it an embedding of $D_4$ into $S_4$, take any bijection $f\colon D_4/H\to\{1,2,3,4\}$ and, for every $g\in D_4$, calculate the corresponding element of $S_4$ defined by: $$k\mapsto f(\varphi(g)(f^{-1}(k))) \tag1 $$ To see this into operation, let's take $\{1, r,r^2,r^3\}$ as a complete set of cosets representatives and the bijection $f$ defined by $r^{i-1}H\mapsto i$, for $i=1,2,3,4$. Then, for example for $g=rs$, the recipe $(1)$ yields: \begin{alignat}{1} 1\mapsto f(\varphi(rs)(f^{-1}(1))) &= f(\varphi(rs)(H)) \\ &= f(rsH) \\ &= f(rH) \\ &= 2 \end{alignat} \begin{alignat}{1} 2\mapsto f(\varphi(rs)(f^{-1}(2))) &= f(\varphi(rs)(rH)) \\ &= f(rsrH) \\ &= f(sH) \\ &= f(H) \\ &= 1 \end{alignat} \begin{alignat}{1} 3\mapsto f(\varphi(rs)(f^{-1}(3))) &= f(\varphi(rs)(r^2H)) \\ &= f(rsr^2H) \\ &= f(srH) \\ &= f(s^2r^{-1}sH) \\ &= f(r^3sH) \\ &= f(r^3H) \\ &= 4 \end{alignat} \begin{alignat}{1} 4\mapsto f(\varphi(rs)(f^{-1}(4))) &= f(\varphi(rs)(r^3H)) \\ &= f(rsr^3H) \\ &= f(sr^2H) \\ &= f(r^2sH) \\ &= f(r^2H) \\ &= 3 \end{alignat} Therefore, the permutation $(12)(34)$ is "$rs$ in $S_4$". And likewise for the other elements of $D_4$. Different choices of the (arbitrary) bijection $f$ lead to set-wise different images of $D_4$ in $S_4$.

(Btw, this routine works verbatim for every $D_n$, as long as $r^2\ne1$, namely for $n>2$.)

Answer 3

[I write $D_{2n}$ for what you call $D_n$.]

I gave one way to generalize the ideas in the comments, to subgroups of $S_k$ that aren't necessarily dihedral. However, dihedral groups are very special: they are precisely those groups generated by two elements of order $2$. If $D_{2n}=\langle a,b\rangle$, with $|a|=|b|=2$, then $n=|ab|$.

In your case, you can take $a=(1234)(24)=(14)(23)$ and $b=(24)$. These both have order $2$, so $\langle(1234),(24)\rangle=\langle a,b\rangle$ is dihedral. Since $ab=(1234)$ has order $4$, we see it is the dihedral group $D_8$, as desired.

Answer 4

Consider the action of $D_4$ on the vertices of a square in the usual way. This induces a homomorphism $D_4\to S_4$. None of the elements act as the identity, as at least one vertex is moved under any symmetry. So, this is an injection. Labelling the vertices $1$ to $4$ tells you the image, which is a subgroup isomorphic to $D_4$.

Hope this helps. :)