1. (a) Find the order of in .
(b) List the elements of the cyclic subgroup of generated by .
(c) Compute the product in the group . (Multiply permutations right to left, and write the result in disjoint cycle form.)
(a) The order of 20 in is . The order of 44 in is . Therefore, the order of in is .
2. List the elements of the cyclic subgroup in .
3. Find a specific element of order 10 in .
Note that . I'll find an element of order 2 in and an element of order 5 in .
There one element of order 2 in , namely 4.
There are several elements of order 5 in ; probably the simplest is 7.
Hence, is an element of order 10 in .
4. is a group under addition. Define by
(a) Prove that f is a group map.
(b) Prove that
Suppose . Then
This proves that .
5. Find an element of having the largest possible order. Explain why no element can have larger order.
If , then . So every element of has order less than or equal to 30.
The element has order 30 in . Therefore, it's an element of the largest possible order.
6. Consider the following subset of :
Either prove that H is a subgroup of , or give a specific counterexample to one of the subgroup axioms.
H is not a subgroup.
H is not closed: , but
Alternatively, , but , contradicting closure under inverses.
The way of progress is neither swift nor easy. - Marie Curie
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