# Show that $S_4/V$ is isomorphic to $S_3$, where $V$ is the Klein Four Group.

(i) Show that $S_4/V$ is isomorphic to $S_3$, where $V$ is the Klein Four Group.

I understand isomorphism to be a bijective homomorphism but I'm unsure how one would go about proving this. $S_4/V$ has order 6 and I think that will be of use but that's as far as I can go.

(ii) Let $G = D_7$ be the dihedral group of order 14 and let $H = C_7$ be the cyclic group of order 7. Find all the homomorphisms from $G$ to $H$.

Are there any homomorpisms at all?

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After these, I think I'll call it a night. Enough maths for today! –  user61854 Feb 14 '13 at 2:39
Your first question is a duplicate of an earlier one. I am uncertain whether we should close this question. The second question saves the day in a way, but it is also a separate question. –  Jyrki Lahtonen Feb 14 '13 at 12:20

Let me add what is possibly the trivial solution to (i).

There is a distinguished copy of $S_3$ in $S_4$, given by the stabilizer of $4$ $$S_3 = \{1, (123), (132), (12), (13), (23) \}.$$ This clearly intersects trivially $$V = \{1, (12)(34), (13)(24), (14)(23) \}.$$ Since $S_3 \cap V = \{1\}$, we have $S_4 = S_3 V$. (From a mildly higher point of view, $V$ is a transitive subgroup, so $S_4 = S_3 V$, as $S_3$ is a one-point stabilizer, and $S_{3}\cap V = \{1\}$ as $V$ acts regularly.)

Now one of the isomorphism theorems tells you that $$S_4/V = S_3 V / V \cong S_{3}/(S_{3}\cap V) = S_3/\{1\} \cong S_3.$$

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I was advised to do something like this, viewing S3 as the subgroup in S4 which fixes the letter 4. Then use these elements as representatives of the cosets to give a bijection and hence show isomorphism. Have you just used one of the elements as a representative? Would I have to do it for the others as well? What are the other elements? (123), (132), (213), (213), (312), (321), (12), (13), (23)? –  user61854 Feb 14 '13 at 15:34
@user61854, be careful, there are only $6$ elements here. You have $(123) = (231) = (312)$ and $(213) = (132) = (321)$. The representatives are exactly the $6$ elements of $S_3$ I listed in my answer. Anyway, my solution is (by chance) exactly the one that was required of you. –  Andreas Caranti Feb 14 '13 at 15:56
actually, not 213 etc. Just the other elements in S3 –  user61854 Feb 14 '13 at 16:01
Ah yes, but what about the other elements? Is the fact they intersect enough? I have only been lectured in one isomorhpism theorem, and I think I can only use that one (&: G/K -> Im(&)). I was told to show a bijection so I don't know if I actually need to use an isomorphism theorem. –  user61854 Feb 14 '13 at 16:05
Thank you, though. You've given me a good insight! –  user61854 Feb 14 '13 at 16:06

Just a “visual” construction of the isomorphism between $\mathfrak S(4)/V_4$ and $\mathfrak S(3)$...

It's quite well known that $4 = 2 +2$. Concretely, it means that if you have four things, you can always group them into two pairs. What's remarkable is that there are exactly three such groupings: $$\{1, 2, 3, 4\} = \{1, 2\} \sqcup \{3,4\} = \{1, 3\} \sqcup \{2,4\} = \{1, 4\} \sqcup \{2,3\}.$$

If I call these three groupings $\mathbf{Order}$, $\mathbf{Parity}$ and $\mathbf{5sum}$, every permutation of $\{1,2,3,4\}$ induces a permutation of these three things. For example, the $4$-cycle $(1234)$ induces the transposition $(\mathbf{Order}\ \mathbf{5sum})$. So I get a homomorphism $$\mathfrak S(4) \to \mathfrak S(\{\mathbf{Order}, \mathbf{Parity}, \mathbf{5sum}\})\simeq \mathfrak S(3).$$ It's quite easy to show that this morphism is surjective (we already found the transposition $(\mathbf{Order}\ \mathbf{5sum})$ in its image, it's not hard to find another transposition or a $3$-cycle, and that will be enough to generate the whole of $\mathfrak S(3)$.)

Let's give a proof that the kernel is exactly $V_4$: let $\sigma$ be a nontrivial permutation in it. So for example, there are two elements $a \neq b$ such that $\sigma(a) = b$ (let's call $c$ and $d$ the two remaining elements) In that case, because $\sigma$ preserves the grouping $\{a,b\} \sqcup \{c,d\}$, it must send $b$ back to $a$. And because it preserves $\{a,c\} \sqcup \{b,d\}$ and it exchanges $a$ and $b$, it must exchange $c$ and $d$ as well. So $\sigma = (a\, b)(c\, d) \in V_4$.

Therefore, the factorisation theorem gives you an isomorphism $\mathfrak S(4)/V_4 \to \mathfrak S(3)$.

Of course, this proof is not the shortest (well, it certainly isn't the shortest to write, but if you're allowed to make a lot of pictures or to play with four actual tokens, essentially all the arguments become self-evident). But it really makes the isomorphism concrete. There's an isomorphism between $\mathfrak S(4)/V_4$ and $\mathfrak S(3)$ because $4$ equals $2+2$ in $3$ different ways...

A final remark: if $n \neq 4$, the only proper normal subgroup of $\mathfrak S(n)$ is $\mathfrak A(n)$. That means that the situation I described here is pretty exceptional. If you want $\mathfrak S(n)$ to act on fewer than $n$ objects, the only interesting actions are :

• for $n$ arbitrary, $\mathfrak S(n)$ acts on two tokens with the following rule: even permutations do nothing, odd ones swap the two tokens (you have to admit this action is pretty boring).
• one exceptional case: $\mathfrak S(4)$ acts on three tokens, as we just saw.

I may be overenthusiastic, but this simple remark gives much cachet to this innocent-looking isomorphism. (Another peculiarity of the same kind is that the only interesting action of $\mathfrak S(n)$ on $n+1$ tokens is an action of $\mathfrak S(5)$ on 6 tokens, coming from another exceptional behaviour of the symmetric group.)

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Hints (complete the argumentation):

(i) $\,|S_4|=4!=24\,\,,\,\,|V|=4\,$ and $\,V\triangleleft S_4\,$ , so $\,\left|S_4/V\right|=6=|S_3|\,$ . Now just prove that there can't be an element $\,\sigma\in S_4\,\,\,s.t.\,\,\,ord_{S_4/V}(\sigma V)=6\,$ (further hint: otherwise, what'd be $\,ord_{S_4}(\sigma)\,$ ?)

(ii) By definition

$$D_7:=\langle r,s\;\;;\;\;s^2=r^7=1\;\;,\;\;srs=r^{-1}=r^6\rangle$$

Now remember that for any homomorphism $\,\phi: D_7\to C_7\;\;,\;\;\phi(D_7)\le C_7\,$ (Further hint: %$\,7\,$ is a prime number...)

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I remember hearing something to do with cyclic groups and prime orders - but I've fogotten :(. How is that of use? –  user61854 Feb 14 '13 at 3:02
Well, the only possible subgroups of $\,C_7\,$ are the trivial ones: $\,\{1\}\,\,,\,\,C_7\,$ , so the image of any homomorphism from any group to $\,C_7\,$ must be one of these two guys...Now remember that you can determine completely any homomorphism from $\,D_7\,$ to whatever by determining the images of its generators... –  DonAntonio Feb 14 '13 at 3:08
I really am not sure how to simplify it anymore: the homomorphic image of $\,D_7\,$ within $\,C_7\,$ is either $\,\{1\}\,$, and this is the trivial homomorphism, or else it is $\,C_7\,$ itself, and this can be accomplished by mapping $\,r\,$ to a generator of $\,C_7\,$ and $\,s\,$ to the unit there, as the order of a homomorphic image of an element divides the elements order. All this is rather elementary stuff and perhaps you should mull on checking your basics in group theory in order to feel more sure in these subject. –  DonAntonio Feb 14 '13 at 3:59
That's part ii), I meant the isomorphism of part i). –  user61854 Feb 14 '13 at 4:02
Read the middle part of my past comment: an element of order six in the quotient would imply an element of order a multiple of six in $\,S_4\,$ ... –  DonAntonio Feb 14 '13 at 4:32

For the first one, note that there are precisely $2$ groups of order $6$ (up to isomorphism)--namely, $C_6$ and $S_3$. All you have to do is show that $S_4/V$ isn't abelian (or, perhaps more simply, isn't cyclic).

For the second, note that $D_7=\langle x,y\mid x^7=y^2=xyxy=1\rangle$, that a homomorphism is completely determined by where it maps a group's generators, and that if $\phi:G\to H$ is a homomorphism, then the order of $\phi(g)$ divides the order of $g$ for each $g\in G$. This should be enough to let you completely determine the homomorphisms $D_7\to C_7$.

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Could I use a cayley table to show part i? –  user61854 Feb 14 '13 at 3:01
You certainly could. –  Cameron Buie Feb 14 '13 at 3:11
Is there a better way? –  user61854 Feb 14 '13 at 3:16
None that come to mind at this hour. Fortunately, you're only dealing with a $6\times 6$ table, and $11$ of the $36$ entries are immediately determined (identity row and column), so it isn't too much work. –  Cameron Buie Feb 14 '13 at 3:24
Would the elements within the table be permutations? I've never seen a Cayley's table with permutations in them before! –  user61854 Feb 14 '13 at 3:34

We see that $S_4/V$ cannot be abelian, because otherwise $S_4' = A_4 \leq V$ which is definitely not true. Thus $S_4/V$ must be nonabelian of order $6$, which means that $S_4/V \cong S_3$.

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Is it as simple as that? Would that be sufficient? I was advised to view S3 as the subgroup in S4 that fixes one letter. Then use these elements as representatives for the cosets giving a bijection but I have no idea how to do so. –  user61854 Feb 14 '13 at 14:49