# Algebraic Structures Question

I am having problems understanding what this question is asking. any help would be appreciated. Thanks.

The dihedral group D8 is an 8 -element subgroup of the 24 -element symmetric group S4 . Write down all left and right cosets of D8 in S4 and draw conclusions regarding normality of D8 in S4 . According to your result determine NS4 (D8) .

• Hint: Write down the elements of $S_4$ using your favorite notation for permutations, in this case on $\{1,2,3,4\}$. Find a subgroup which represents $D_8$. Now it makes sense to talk about writing down cosets of $D_8$ inside $S_4$. Nov 23 '11 at 4:05
• It's hard to give useful advice when you don't tell us where you are stuck. Do you not know what $S_4$ is? Do you know what $S_4$ is, but not what $D_8$ is? Do you know both of these, but you don't see how $D_8$ could be a subgroup of $S_4$? Do you understand that much, but you don't know what left and right cosets are? Do you know about cosets, but you don't know what it means for one group to be normal in another? Nov 23 '11 at 4:27
• Yes, I understand that if the left cosets equal the right cosets then the group is normal and i understand that to get the left and right cosets you have to multiply the elements in D8 by the elements in S4 i don't understand however, how you get these element for either D8 or S4. Thank you for your help. Nov 23 '11 at 5:00
• The elements of $S_4$ are just all permutations of $\{1,2,3,4\}$; the identity, the six two-cycles, the eight 3-cycles, the six 4-cycles, and the three products of two 2-cycles. To view $D_8$ as a subgroup, take your definition of $D_8$ (it was either given as permutations already, or as rigid motions of the square; in the latter case, number the vertices, and see how each element of $D_8$ permutes them). Nov 23 '11 at 5:27
• Thank you, so does this mean that effectively to get the right cosets you have to multiply D8 by 22 elements i.e. the elements of S4 and that the elements of D8 are {id, (2341), (3412), (4123), (2143), (4321), (1432), (3214)} Nov 23 '11 at 5:50

Represent $D_8$ with your preferred notation. Perhaps it is the group generated by $(1234)$ and $(13)$. That's ok. Then write down the 8 elements. Then multiply each on the right and on the left by elements of $S_4$, i.e. write down the right and left cosets. You can just sort of do it, and I recommend it in order to get a feel for the group. Patterns will quickly emerge, and it's not that much work.
• @Sarah - hmm. Well, $D_8$ is the set of isometries of a square. So it's the 4 rotations and 4 reflections (I counted the trivial rotation) that make the square look at if it hadn't moved, but where the vertices do change (so we label the vertices and pay attention to how they change). $S_4$ is the group of permutations of 4 elements. Some include $(123)$, which sends 1 to 2, 2 to 3, and 3 to 1. Or $(12)(34)$, which sends 1 to 2, 2 to 1, 3 to 4, and 4 to 3 (so you read each set of parenthesis as a cycle). Nov 23 '11 at 7:57