information about semi-dihedral groups. my question is about the elements and the generalized format of caylay table of groups called semi-dihedral groups which have the presentation $$ \langle a,b\mid a^{4m}=b^2=1,ab=ba^{2m-1}\rangle $$


*

*what is the elements are?

*what is the format of caylay table?like we know that there is a beautiful format for dihedral groups.

*what is the subgroups of these kind of groups?


if there is a reference that I can study,it will be great if you name it, and also any hint or guidance will be nice. thanks a lot.
I don't know maybe finding other properties will be easier, I want to know do these kind of groups have cyclic sylow 2-subgroups?
if $m=2^{\alpha}$ then sylow 2-subgroups is itself and it is not cyclic,for $m=3,5,6$ I have checked sylow 2-subgroups with GAP and they are not cyclic,I guess that it is true for all,but I need to have it's subgroup structure!and I don't know it.
if I show that semidihedral groups have no normal subgroups of odd order,again I have that there is no cyclic sylow 2-subgroups.but I don't know how to show that.
it will be great if you share your Ideas with me.thanks.
 A: This presentation gives you somewhat more concrete  definition of whatever group you have with generators a,b and relations given by the one that you posted. Well you know that those are your generators so one way to form your cayley table is just use the generators and your relations to form the elements multiplication. For example we have dihedral group have the following presentation 
$D_{2n}$ = $<r,s | r^n = s^2 = 1, sr = rs^-1>$, so from this we can already see that r and s are subgroups of your group G, however the problem with presentations is sometimes there is some relation that could collapse the group into something that is trivial for example I was solving this problem earlier and the idea of it is that we try and mimic the group presentation of $D_2n$ to get something similar for example
$X_2n$ = $< x,y | x^n = y^2 = 1, xy = yx^2>$   and we would expect that this group would have similar order like dihedral group that is 2n however if we have n = 2k then we can prove that ord(G) = 6 regardless for what k we can choose ! I have added a link to my question maybe it will help you out more !
