I understand that the order of a subgroup (if it exists) must divide the order of the group so subgroups (if they exist) of D4 (order 8) must be of order 4, 2 and 1.
But how do we find these subgroups - apart from trial and error followed by a normal subgroup test - or is that the only way?
I am looking for a smarter approach that is better than a trial an error checking of all potential subgroups (of the right orders) followed by a check of normality. But perhaps there is no smarter way.
Peter
any subgroup of order 4 (should we find some) has to be normal. why? because such a subgroup (let's call it H for now), has only two right cosets, H and Hx, for some x not in H.
since x is not in H, the left coset xH must be the same as Hx, so in turn xHx^-1 = H, and H is normal.
there is only ONE subgroup of order 1, and only ONE subgroup of order 8, both of which are necessarily normal, but not very interesting.
there are (up to isomorphism) 2 groups of order 4, the cyclic group of order 4, and the klein 4-group.
the former has 2 elements of order 4 and 1 of order 2, and the latter has 3 elements of order 2. both are abelian.
so...we should look first for elements of order 4, to see if D4 has any cyclic subgroups of order 4.
let's list the elements of D4, first, just for reference (your text may use different notation):
D4 = {e,r,r^2,r^3,s,rs,r^2s,r^3s}, and we have the multiplication rules r^4 = s^2 = e; sr = r^3s.
examining the element orders, we have just 2 elements of order 4: r and r^3. it is easy to see that <r> = <r^3> = {e,r,r^2,r^3}.
this is one subgroup of order 4, and it has to be normal. it is also clearly the only cyclic subgroup of D4 of order 4.
now, if we want to find a group of order 4 isomorphic to the klein 4-group, we need 3 elements of order 2,
and since the klein 4-group is abelian, these all have to commute with each other. here are the elements of order 2 we have to choose from:
{r^2, s, rs, r^2s, r^3s}. there are 5!/(3!2!) = 120/12 = 10 ways to choose 3 elements, which is a lot of trial-and-error :S
well, suppose our subgroup contained s. sr^2 = (sr)r = (r^3s)r = r^3(sr) = r^3(r^3s) = r^6s = r^2s, so r^2 and s commute.
s(rs) = (sr)s = (r^3s)s = r^3 ≠ r = (rs)s, so if our subgroup contains s, it could also contain r^2, but it could not contain rs.
it should be clear that since s commutes with r^2, it also commutes with r^2s, and closure is obvious,
so wham! {e,r^2,s,r^2s} is a 2nd subgroup of order 4.
since s(r^3s) = (sr)r^2s = (r^3s)r^2s (and we know s commutes with r^2s)
= r^3(r^2)s^2 = r^5 = r, but (r^3s)s = r^3, so this is the ONLY possible subgroup of order 4 containing s (because s only commutes with r^2 and r^2s).
since 6 of our 10 choices of 3-element subsets of {r^2,s,rs,r^2s,r^3s} contain s, that leaves just 4 choices of 3 elements of order 2 left:
{r^2, rs, r^2s}, {r^2, rs, r^3s}, {r^2, r^2s, r^3s} and {rs, r^2s, r^3s}.
we can eliminate the last one straight-away, since rs(r^2s) = r(sr)(rs) = r(r^3s)(rs) = r^4s(rs) = srs = r^3s^2 = r^3,
so we don't even have closure. 3 to go....
rs(r^2s) = r(sr)(rs) = r(r^3s)(rs) = srs = ....this is familar.....again, closure fails. you prove the other 2 left don't work.
so...that leaves just the subgroups of order 2. these are easy, any element x of order 2 generates a subgroup of order 2 {e,x}.
you should be able to find 5 such subgroups. only one of them will turn out to be normal (just compute left and right cosets until
you find some coset xH not equal to Hx, for each subgroup H).
Thanks so much for such a fulsome answer ... really appreciate your help
Will have to go and work through this.
Just a quick question straight away. You write "any subgroup of order 4 (should we find some) has to be normal. why? because such a subgroup (let's call it H for now), has only two right cosets, H and Hx, for some x not in H. Since x is not in H, the left coset xH must be the same as Hx, so in turn xHx^-1 = H, and H is normal."
How do you know H has only two cosets? And how does the second fact follow?
Again, thanks!!
Peter
suppse the order of H is 4. then any coset has 4 elements. cosets are either identical, or distinct, they never "partially overlap".
suppose Hx ∩ Hy is non-empty: so we have z in Hx, and z in Hy.
this means z = hx, for some h in H, and z = h'y for some h' in H, so hx = h'y. this tells us 2 things:
x = h^-1h'y, so x is in Hy, so Hx is a subset of Hy.
y = (h')^-1hx, so y is in Hx, so Hy is a subset of Hx. since the 2 cosets contain each other, they are equal.
so if 2 cosets intersect at all, they intersect "all the way". otherwise, they don't intersect at all.
so if H has 4 elements, the remaining 4 elements of D4 have to be "the other coset", which is Hx or xH, for some (any) x that isn't already in H.
for example, if H = <r> = <e,r,r^2,r^3>, the two cosets are H and Hs = {s,rs,r^2s,r^3s}.
calculating sH, we get: {s,sr,sr^2,sr^3}, and sr = r^3s, while sr^3 = (sr)r^2 = (r^3s)r^2 = (r^3)(sr)r
= (r^3)(r^3s)r = r^6(sr) = r^2(sr) = r^2(r^3s) = r^5s = rs, so sH = {s,r^3s,r^2s,rs} = Hs (in a slightly different order, but same elements).
now...how does xH = Hx imply xHx^-1 = H?
if xH = Hx, that means that for every element in xH, xh, we can write it as h'x, for some (perhaps different) h' in H.
so xh = h'x --> xhx^-1 = h', so xHx^-1 is contained in H. similarly, given any hx in Hx, we can write:
hx = xh" for some h" in H. so h = xh"x^-1, and H is contained in xHx^-1, so...H = xHx^-1.
so if EVERY left coset xH equals it's matching right coset Hx, then xHx^-1 = H for all x in G, so H is normal.
when we have only 2 cosets to work with, this is easy to see.
we can summarize this as if |G|/|H| = 2, H is normal in G.