Let $H = \{\text{id}, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3) \} \subset S_4$, and let $ K = \{\sigma \in S_4 \mid \sigma(4) = 4\}.$
(a) How to show that $H$ is a subgroup of $S_4$, and is $H$ is a normal subgroup? What other group is $H$ isomorphic to? Same for $K.$
(b) Does every coset of $H$ contain exactly one element of $K.$ Also, can every
element of $S_4$ be written uniquely as the product of an element of $H$ and an element of $K$?
(c) What can be said about the quotient group $S_4/H$? Is $S_4$ isomorphic to the direct product $H × K$?
For part a, to show that H is a subgroup of S4, I will take each element in S4 and conjugate it by what? i.e., to be normal NH = HN, or g-1ng = n
for part b, I am not sure what is being asked of
Best Answer
To show that $H$ is a subgroup of $S_4$, show that every element of $H$ is in $S_4$ (that is, that $H\subseteq S_4$); that $H$ is not empty (a given); that the product of any two elements of $H$ is again in $H$; and that the inverse of any element of $H$ is again in $H$. The only difficult part will be showing the product. Multiplying by the identity is easy, so you just need to worry about the other products.
To show that $H$ is normal, you need to show that for every $\sigma\in S_4$ and every $h\in H$, $\sigma \circ h \circ\sigma^{-1}\in H$. Do you know something about the "shape" (type of disjoint cycle decomposition) of a permutation of the form $\sigma\circ\tau\circ\sigma^{-1}$, in terms of the "shape" of $\tau$? If you do, then it's easy to do in this case.
$H$ has four elements. There aren't many groups with four elements (in fact, there's exactly two nonisomorphic groups with four elements). Which of them is isomorphic to $H$?
Similarly with $K$; though I would be careful in trying to figure out whether $K$ is normal or not. How many elements does $K$ have? Convince yourself that it has $6$ elements. Again, there aren't many groups with six elements (in fact, there's exactly two nonisomorphic groups with six elements). Which one of them is isomorphic to $K$?
$H$ has exactly six cosets in $S_4$. Each of them has four elements. Now, $K$ has six elements. The first question is just: do the elements of $K$ lie each in a different coset, or is there a coset that contains more than one element from $K$?
For the second question: $HK= \{hk \mid h\in H, k\in K\}$. The question is: does the set $HK$ equal $S_4$?
Alternatively: remember that the cosets are sets of the from $$H\sigma = \{h\sigma\mid h\in H\}.$$ Since the cosets form a partition of $S_4$, if you pick six elements of $S_4$, $\sigma_1,\ldots,\sigma_6$ such that $S_4 = H\sigma_1\cup\cdots\cup H\sigma_6$, then every element of $G$ can be written as $h\sigma_i$ for some $h\in H$ and some $i\in\{1,2,\ldots,6\}$. Can you pick $\sigma_1,\ldots,\sigma_6$ so that they are all elements of $K$
Let us take for granted that the answer in part 2 is "yes", the elements of $K$ are distributed among the six cosets, and every element of $G$ can be written as $hk$ with $h\in H$ and $k\in K$. Prove that this is unique (that is, there is only one way of doing it). Now consider the map $G\to K$ given by sending $hk$ to $k$. Is it a homomorphism? (Hint: $(hk)(h'k') = h(kh')(k^{-1}k)k' = \Bigl( h(kh'k^{-1})\Bigr)(kk')$). What is the kernel.
Now, if $S_4$ is isomorphic to $H\times K$, then you would have a subgroup of $S_4$ with $4$ elements, and a subgroup of $S_4$ with six elements (corresponding to the subgroups $H\times\{\mathrm{id}\}$ and ${\mathrm{id}}\times K$) that intersect at the identity, and where every element of the first one commutes with every element of the second one. Since groups of order $4$ are abelian, that would mean that the center of $S_4$ has at least four elements. What is the center of $S_4$?