Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
James, it is not considered polite here to just paste in a question and wait for answers - it does not show that you have thought about the problem. Please explain what you've tried so far, and where you are stuck. – Zev Chonoles Mar 6 '12 at 10:35
Try drawing a picture of a square and labeling the vertices cyclically. How does H act on your picture? – Louis Mar 6 '12 at 10:39
What methods do you know for showing something is a subgroup? for showing something is a normal subgroup? Do you know all the groups with the same order as $H$, and how to tell them apart? Can you list the elements of $K$? Can you list the cosets of $H$? – Gerry Myerson Mar 6 '12 at 11:30
@Zev, I did explain some stuff but I need help in understanding the question – James R. Mar 6 '12 at 12:35
I think Gerry Myerson shouted the answer too loud that I am hearing it right away. I would add an answer if nobody else does after a day. I hope that would be long enough or may be not. – user21436 Mar 6 '12 at 12:53
  1. 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 $K$?

  2. $H$ 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 $K$

  3. 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 $S_4$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.