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November 3rd, 2012, 01:05 AM   #1
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Isomorphism

I was recently revising abstract algebra and doing some exercises. I got stuck on this one :

let where G is the dihedral group of order 16. It says to compute the center of and determine the isomorphism type of .

I conclude that Hence, G' has order 4 since is being partitioned into pairs.

Since it has order 4, G' is either isomorphic to Z4 or V4. Now how can I specify G' is isomorphic to which one of these two?

Balarka

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November 4th, 2012, 02:16 PM   #2
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Re: Isomorphism

it turns out that <r^4> = Z(G), so G' = (G/Z(G))/Z(G/Z(G)).

since r has order 8 in G, r^4 has order 2, so |G/<r^4>| = 8.

before we even look at Z(G/<r^4>), we might want to figure out which elements of G yield distinct cosets of G/<r^4>. for notational reasons, i'll just call <r^4> = Z.

we have as elements of G/Z:

Z = {1,r^4}
rZ = r^5Z = {r,r^5}
r^2Z = r^6Z = {r^2,r^6}
r^3Z = r^7Z = {r^3,r^7}
sZ = sr^4Z = {s,sr^4}
srZ = sr^5Z = {sr,sr^5}
sr^2Z = sr^6Z = {sr^2,sr^6}
sr^3Z = sr^7Z = {sr^3,sr^7}

if i write [x] for xZ (i don't feel like fussing with latex and "overline" notation), we have:

G/Z = {[1],[r],[r^2],[r^3],[s],[sr],[sr^2],[sr^3]}

which group of order 8 is this? first, let's check the order of [r]:

[r]*[r] = [r^2] <--not of order 2 (which leaves 4 and 8 to check).

[r^2]*[r^2] = [r^4] = [1] <--- [r] has order 4.

note that [s]*[s] = [s^2] = [1] <--- [s] has order 2.

is G/Z abelian? it appears to be generated by [r] and [s], so let's check [r]*[s] and [s]*[r].

[r]*[s] = [rs] = [sr^7] (because rs = sr^7 in G)

= [sr^3]

[s]*[r] = [sr] ? [sr^3], so G/Z is NOT abelian (and we don't have to worry about checking for elements of order 8. there aren't any, since cyclic groups are abelian).

IF [r]*[s] = [s]*([r]^-1) = [s]*[r^3] = [s]*([r]^3), we would know that G/Z is isomorphic to the dihedral group of order 8.

and indeed, we just showed that above, since [sr^3] = [s]*[r^3] = [s]*([r])^3.

but you should know (hopefully) that Z(D4) = {1,r^2}.

so Z(G/Z) = {[1],[r^2]}. let's call THIS subgroup of Z/G, H (or we'll be overwhelmed with brackets and parentheses).

here are the cosets of H:

H = {[1],[r^2]}
[r]H = {[r],[r^3]}
[s]H = {[s],[sr^2]}
[sr]H = {[sr],[sr^3]}.

now if (G/Z)/H is cyclic, it has 2 elements of order 4. so it suffices to check the orders of 2 non-identity elements, if our group is cyclic, one of those 2 has to be of order 4.

([r]H)*([r]H) = ([r])^2H = [r^2]H = H <---[r]H has order 2.
([s]H)*([s]H) = ([s])^2H = [s^2]H = [1]H = H <---[s]H has order 2.

we therefore conclude that G' is isomorphic to V.
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November 4th, 2012, 11:53 PM   #3
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Re: Isomorphism

Wow, Thanks Deveno!
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