MTH 236H: Series of Groups

Luke Cybulski

April 18, 2014

Problem 35.2The first series has factor groups of order 20, 3, and 60Z/{0} is simply isomor-phic to 60Z, and so is infinite order. Next, the second series has factor groupsof order 49, 5, and 245Z/{0} is isomorphic to 245Z, and so is infinite order. So,from these, we need to add factor groups of order 49 and 5 to the first series andfactor groups of order 20 and 3 to the second series. This can be done by in-serting factor groups before 60Z and 245Z. Thus, the isomorphic refinementsare:

{0} < 14700Z < 300Z < 60Z < 20Z < Z{0} < 14700Z < 735Z < 245Z < 49Z < Z

Problem 35.4The first series has group orders of 1, 4, 24, and 72 (read from left to right).The second series has group orders of 1, 3, 6, and 72. So, we can use Lagrangestheorem to see that the first series has factor groups of order 3, 6, and 4, andthat the second series has factor groups of order 12, 2, and 3. So, for the firstseries, we can insert 36 between {0} and 18, and this gives factor groups oforder 3, 6, 2, and 2. Next, we insert 6 in between 12 and Z72, and this givesfactor groups of order 6, 2, 2, and 3. Thus, we see this gives the isomorphicrefinements of the two series:

{0} < 36 < 18 < 3 < Z72{0} < 24 < 12 < 6 < Z72

Problem 35.5The first series has factor groups of order 10, 6, and then (60Z Z)/{(0, 0)} isisomorphic to 60ZZ, and is thus of infinite order. The second series has factorgroups of order 20, 4, and then (Z 80Z)/{(0, 0)} is isomorphic to Z 80Z,and is thus of infinite order. So, for isomorphic refinements, we must add factorgroups of order 20 and 4 into the first series and then factor groups of order10 and 6 into the second series. This can be done by inserting factor groupsbefore 60Z Z and Z 80Z. Thus, the isomorphic refinements are:

{(0, 0)} < 4800Z Z < 240Z Z < 60Z Z < 10Z Z < Z Z{(0, 0)} < Z 4800Z < Z 480Z < Z 80Z < Z 20Z < Z Z

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Problem 35.12First, the center of S3 is simply the identity {0}. Next, the center of D4 is theset {0, 2}. Thus, the center of the direct product S3 D4 is simply given by:

Z(S3 D4) = Z(S3) Z(D4) = {0} {0, 2} = {{0, 0}, {0, 2}}

Problem 35.13First, the center of S3 is simply the identity {0}. Next, it is clear that thecenter of Z4 is Z(Z4) = Z4, considering Z4 is abelian. Thus, the center of thedirect product S3 Z4 is given by:

Z1(S3 Z4) = Z(S3) Z(Z4) = {0} Z4Now, we consider the factor group (S3 Z4)/Z1(S3 Z4).

(S3 Z4)/Z1(S3 Z4) = (S3 Z4)/({0} Z4) = S3We then know that the next element of the series, Z2, can be defined inductivelyso that Z2/Z1 = Z((S3Z4)/Z1). Looking at the above factor group, it is clearthat Z((S3 Z4)/Z1) = {{0} Z4}. Thus,

Z2/Z1 = Z2/({0} Z4) = {{0} Z4}

This gives that Z2 = {0} Z4 (Recall that for some group G, G/G = {e}.In our case, {0} Z4 is the identity element of the factor group, so it mustbe that Z2 = Z1, else we would not get Z2/Z1 = {e}). Then, we get Z3from Z3/Z2 = Z((S3 Z4)/Z2), but since Z2 = Z1 and Z((S3 Z4)/Z2) =Z((S3 Z4)/Z1) = {{0} Z4}, it is clear that Z3 = Z2. So, continuing thisprocess gives Zi = {0} Z4. Then, we know the ascending central series isgiven by

Z1 Z2 Z3 Thus, the ascending central series for S3 Z4 is

{0} Z4 {0} Z4 {0} Z4

Problem 35.16This definition is incorrect, i.e., it says a composition series of abelian groups,while it should say a composition series such that all factor groups Hi+1/Hi areabelian. Therefore the correct definition is technically:

A group G is solvable if it has a composition series {Hi} such that all factorgroups Hi+1/Hi are abelian.

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Problem 35.17 (proofs are given for non-trivial answers)

(a):True(b):False(c):True(d):False(e):False(f):True

First, if G is simple, we use the composition series {e} < G. Next, if theorder of G is 1, then we are simply left with the trivial composition series. So,we then have the last case, which is that G is finite and that G is not simple, i.e. a proper nontrivial normal subgroup of G. Furthermore, we know there willbe at least one maximal normal subgroup, say M1. Then, we know that G/M1is simple. Thus,we can form the composition series {e} < M1 < G. If M has atleast proper non-trivial normal subgroup, one will again be a maximal normalsubgroup, say M2. Thus, M1/M2 is simple and we have the composition series{e} < M2 < M1 < G. We can continue to use this same process, but sinceG is finite, it will eventually stop, i.e., the only normal subgroup left of somemaximal normal subgroup Mi will just be {e}. So, the final composition serieswill be {e} < Mi < . . . < M2 < M1 < G.(g):False(h):False

First, we note that the factor group Sn/An = Z2, and thus simple. Next,An/{e} = An, and for n = 2, 3 and n 5, An is simple. So, the compositionseries for S7: {e} < A7 < S7. Yet, A7 is not abelian, thus S7 is not solvable.(i):True(j):True

If |G| = p for p prime, then by Lagranges theorem, G can only have theimproper subgroup and the trivial subgroup. Thus, G is clearly simple, and so the composition series {e} < G. Next, all groups of prime order are cyclic,hence G is cyclic. Moreover, we know all cyclic groups are abelian, thus thefactor group G/{e} = G is abelian, and G is solvable.

Problem 35.18 Let e be the identity element in S3. Now, we first see that thefactor group (S3S3)/(S3{e}) = S3, but S3 is not simple. Thus, starting withthe series {e}{e} < S3{e} < S3S3 we can refine it by inserting S3A3 inbetween S3 S3 and S3 {e}. Furthermore, (S3 {e})/({e} {e} = S3 {e},which is not simple, so we insert A3 {e} in between S3 {e} and {e} {e}.This gives the following refined series:

{e} {e} < A3 {e} < S3 {e} < S3 A3 < S3 S3From this, we see the factor groups have orders (starting from (S3S3)/(S3A3)2, 3, 2, and 3. Thus, since these orders are prime, we know that all the factorgroups of simple. Thus, the prior series is a composition series of S3S3. Next,we know that all groups of order 2 and 3 are abelian, thus S3 S3 is solvable.

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