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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 $\bar{G}= G/=$ where G is the dihedral group of order 16. It says to compute the center of $\bar{G}$ and determine the isomorphism type of $G' = \bar{G} / Z(\bar{G})$. I conclude that $Z(\bar{G})= \{\bar{1}, \bar{r}^4\}$ Hence, G' has order 4 since $\bar{G}$ 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 = {[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.
 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!

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