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 November 3rd, 2012, 01:05 AM #1 Math Team   Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 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 . November 4th, 2012, 02:16 PM #2 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Isomorphism it turns out that = 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/| = 8. before we even look at Z(G/), we might want to figure out which elements of G yield distinct cosets of G/. for notational reasons, i'll just call = 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 = {,[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] =  <--- [r] has order 4. note that [s]*[s] = [s^2] =  <--- [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) = {,[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 = {,[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 = H = H <---[s]H has order 2. we therefore conclude that G' is isomorphic to V. November 4th, 2012, 11:53 PM #3 Math Team   Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Isomorphism Wow, Thanks Deveno! Tags isomorphism Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post tiger4 Abstract Algebra 2 July 2nd, 2012 03:17 AM bewade123 Abstract Algebra 2 February 14th, 2012 05:12 PM jpav Abstract Algebra 6 July 11th, 2011 07:00 AM mia6 Linear Algebra 1 November 10th, 2010 09:31 AM just17b Abstract Algebra 4 December 18th, 2007 08:57 AM

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