Use Lagrange’s Theorem in the multiplicative group to prove Fermat’s Little Theorem: if is prime then .
If is prime, then (where denotes the Euler totient). Thus . So for all , we have divides . Hence mod .
Use Lagrange’s Theorem in the multiplicative group to prove Fermat’s Little Theorem: if is prime then .
If is prime, then (where denotes the Euler totient). Thus . So for all , we have divides . Hence mod .
Fix , and let act on in the natural way. Show that .
It is clear that . Removing from each function in this set the pair , we have the full symmetric group on . We saw in a previous exercise that the isomorphism type of depends only on the cardinality of , so that the conclusion holds.
Prove that does not have a normal subgroup of order 8 or a normal subgroup of order 3.
Let be a subgroup of order 3; then for some . Note that and , so that is not normal in .
Now suppose is a subgroup of order 8. Now the order of every element in must divide 8 by Lagrange, and no element of has order 8. Moreover, has 9 elements of order 2: 6 2-cycles and 3 products of disjoint 2-cycles. Suppose does not contain an element of order 4. If contains all 3 products of 2 cycles, then (WLOG) we have , a contradiction. If does not contain all the products of 2 cycles then it must contain at least 5 2-cycles; WLOG then contains elements , , and , and , a contradiction. Thus contains a 4-cycle.
Say . Then , where . Suppose for a moment that contains another element of order 4; without loss of generality, is one of and . If , then . But then , but has order 3. Thus . Similarly, , hence the remaining elements of all have order 2. There are 8 such elements remaining.
Note that , , , and , so that , , , and are not in . Then the remaining elements must be in , so that , , , , , , . A simple calculation shows that indeed .
We saw in the previous exercise that, if , then . By Corollary 15 in the text, then, is not normal in .
Fix any labeling of the vertices of a square and use this to identify as a subgroup of . Prove that the subgroups and do not commute in .
We can label the vertices of a square as follows.
Now a 90 degree clockwise rotation corresponds to the permutation and a reflection across the (1,3) axis to .
Now let and . If , then in particular , a contradiction. So .
Let be a group and . Prove that the map sends each left coset to a right coset (of ), and hence .
(This is a slightly different approach.)
Define by . Suppose ; then . and we have , so . By a lemma to a previous exercise, this induces a mapping given by . is clearly surjective, so is surjective. Now suppose ; then , so that in particular and hence . Thus . Hence is a bijection.
Let be a group and let be subgroups of finite index; say and . Prove that . Deduce that if and are relatively prime, then .
Lemma 1: Let and be sets, a map, and an equivalence relation on . Suppose that if then for all . Then given by is a function. Moreover, if is surjective, then is surjective, and if implies for all , then is injective. Proof: is clearly well defined. If is surjective, then for every there exists such that . Then , so that is surjective. If , then , so that , and we have .
First we prove the second inequality.
Lemma 2: Let be a group and let be subgroups. Then there exists an injective map . Proof: Define by . Now if , then we have , so that , and , so that . Thus . Moreover, if then we have , so that . By Lemma 1, there exists an injective mapping given by .
As a consequence, if and are finite, .
Now to the first inequality.
Lemma 3: Let be a group and . Let be a set of coset representatives of . Then the mapping given by is bijective. Proof: (Well defined) Suppose . Then , so that , and we have . (Surjective) Let . Now for some ; say . Then , so that is surjective. (Injective) Suppose . Then ; in particular, , so that and hence . So , and in fact . Thus , and is injective.
As a consequence, we have .
Now in this case we have . Thus divides and divides , so that divides . In particular, since all numbers involved are natural, .
Finally, if and are relatively prime, then , and we have .
Let be a finite group and let be a prime dividing . Let denote the set of all -tuples of elements of whose product is 1. That is, . Now define a relation on as follows: if and only if is a cyclic permutation of .
Hence is an equivalence relation.
Trivial.
Let be a group and let be finite subgroups of relatively prime order. Prove that .
Let and . We saw in a previous exercise that is a subgroup of both and ; by Lagrange’s Theorem, then, divides and . Since , then, . Thus .
Let be a group and . Define a relation on by if and only if . Prove that is an equivalence relation and describe for each the equivalence class . Use this to prove the following proposition.
Proposition: Let be a group and . Then (i) the set of left cosets of is a partition of and (ii) for all , if and only if .
So is an equivalence.