Math 546
Problem Set 12
1. (a). Find a 6-element subgroup H of Z 24 and list all the left cosets of H.
Solution: Choose H = {4,!8,!12,!16,!20,!0} .
The cosets are H = {4,!8,!12,!16,!20,!0} , 1 + H = {5,!9,!13,!17,!21,!1} ,
2 + H = {6,!10,!14,!18,!22,!2} , and 3 + H = {7,!11,!15,!19,!23,!3} .
Notice that these four sets form a partition of Z 24 .
(b). Find a 4-element subgroup H of Z 24 and list all the left cosets of H.
Choose H = {6,!12,!18,!0} .
The cosets are H = {6,!12,!18,!0} , 1 + H = {7,!13,!19,!1} , 1 + H = {7,!13,!19,!1} ,
and 1 + H = {7,!13,!19,!1} , 1 + H = {7,!13,!19,!1} , and 1 + H = {7,!13,!19,!1} .
Notice that these six sets form a partition of Z 24 .
(c). Find a 3-element subgroup H of Z 24 and list all the left cosets of H.
Solution: Choose H = {8,!16,!!0} . There are 8 cosets.
(d). Find a 2-element subgroup H of Z 24 and list all the left cosets of H.
Solution: Choose H = {12,!!0} . There are 12 cosets.
2. (a). We know that Z12 is cyclic. Which elements of Z12 generate it?
Solution: the elements that generate Z12 are 1, 5, 7, and 11.
(b). Make a reasonable guess as to which elements of Z n are generators of Z n .
Solution: the generators of Z n are the elements of U(n) .
(c). Verify your guess.
Let m !U(n) . Then since U(n) is a group with identity 1, there exits an inverse
for m in U(n) . Say w !U(n) such that mw = 1 , where the multiplication is
modulo n. Hence, mw ! 1mod n . But now if k is any element of Z n , then
m(kw) ! k mod n . This means that m to the kw power in Z n is equal to k. In other
words every element of Z n is a power of m.
Note: In Z n , raising k to the power 3 is the same as k + k + k = 3k etc.
If m is not an element of U(n) , then n’power’ of m can every yield the number 1
which is also an element of U(n) .
So, how many generators of Z n are there?
3. Show that any two cyclic groups of the same order are isomorphic. This is why
we tend to speak of “the cyclic group of order 6” instead of “a cyclic group of
order 6.”
4. Show that if H is a subgroup of the group G, then all the left cosets of H have the
same cardinality.
Hint: For b in G, what is a natural function from H to bH ?