# Show that $S_4/K \cong S_3$ [duplicate]

Let $K$ be the subgroup of $S_4$ such that $K = \{e, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)\}$.

I'm not sure if the first part of the question is related, but it asks to show that $K \triangleleft S_4$.

I believe I've shown that $K \triangleleft S_4$.

For any $g \in S_4$, $g$ can be expressed as a product of transpositions

The definition of normal means we need to show $gKg^-1 = K$

So suppose we represent $K$ as arbitrary transpositions then: $K = \{e, (a,b)(c,d), (a,c)(b,d), (a,d)(b,c)\}$

Then for an arbitrary transposition in $S_4$:

$$(a,b) \{e, (a,b)(c,d), (a,c)(b,d), (a,d)(b,c)\} (a,b)$$ $$= \{e, (a,b)(c,d), (a,c)(b,d), (a,d)(b,c)\}$$ $$= K$$

So $K$ is normal.

Now I need to show that $S_4/K\cong S_3$.

I'm pretty sure that I have to find a map $\phi:S_4 \rightarrow S_3$ that is a homomorphism such that $\ker(\phi) = K$, but I'm not sure how to do this.

Your proof that $K$ is a normal subgroup of $S_4$ is not quite sufficient. It is enough to show that $\tau K\tau=K$ for any transposition $\tau\in S_4,$ but you haven't justified it. (Of course, you don't have to, if you already know that result.) So, take an arbitrary transposition $(a,b)\in S_4$ and show that $$(a,b)\bigl\{e,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\bigr\}(a,b)=K,$$ which you still haven't done.
Now, you could then finish the proof that $S_4/K\cong S_3$ by finding a homomorphism as you describe. Alternately, note that $S_4/K$ is a group (why?) with $6$ elements (why?), so is either cyclic or isomorphic to $S_3$ (why?).
• For any transposition $\tau \in S_4$, it's inverse is itself so showing $\tau K \tau = K$ shows it is normal because the definition of normal is $\forall n \in N$ and $\forall g \in G$, $gng^{-1} \in N$. Would this be enough to justify it? For the second part, $S_4 / K$ is a group because $K$ is normal so it forms the quotient group. Since the order of $S_4 = 4! = 24$ and order of $K = 4$, then the order of $S_4 / K = 24/4 = 6$ so it has 6 elements. I'm not really sure why it is isomorphic to $S_3$. Is it because $S_4 / K$ has the same order as $S_3$? – user3370201 Dec 10 '15 at 0:06
• That still doesn't quite do it. As a hint for how to finish justifying it, use a decomposition of an arbitrary $g$ into transpositions to find a decomposition of $g^{-1}$ into transpositions. You are absolutely right in your proof that $S_4/K$ is a group with $6$ elements, but that isn't quite enough. However, one can prove that if a group $G$ has order $6,$ then either $G$ is cyclic or $G\cong S_3.$ Using that result, all that remains is to show that $S_4/K$ isn't cyclic. – Cameron Buie Dec 10 '15 at 0:31