Finite abelian groups

Let be a finite abelian group. Then has exactly one -Sylow subgroup, namely Proof:
Since all -Sylow subgroups are conjugate, and conjugations does nothing to an abelian group, there can only be one -Sylow subgroup, call it . Now we must show .

: From Corollary (b)Corollary (b) we know that since is a -group, .

: Firstly we show that is a subgroup. For any there are such that . WLOG, let . Then , meaning , so . Furthermore, we have , so , meaning is a subgroup. By Corollary (b)Corollary (b) we know is a -subgroup, meaning there must exist a -Sylow subgroup containing , and since is the only -Sylow subgroup, .

Let be a finite abelian group and be subgroups of . The homomorphism is injective if and only if for every , .

Proof:
: We know , so being injective means that this is the only product that gives . If there would be some in and , then so would be obtainable by making and , and .

: We want to show that . Assume . Then meaning , meaning . By induction, every .

Let be a finite abelian group. Then is the direct product of its -Sylow subgroups. In other words, if Proof:
The map given by is a homomorphism since is abelian. We first use LemmaLemma to show that is injective. Assume where each . Then there are 's such that , meaning where are exponents we don't care about since they are multiplying anyway. Since we also know , we have that is a common factor of and , but these are all distinct primes, so meaning . Hence is injective. It follows from the sizes of -Sylow groups that the product of the sizes of is , meaning is also surjective. So is indeed an isomorphism between and .

Let be an abelian -group. Let be of maximal order. Then there exists a subgroup such that is an isomorphism.

Proof:
Since is a -group, . We induce on . If then , so the base case holds. Now take any . We know for some .

Claim 1: For every , .
Proof: We know for some , implying If we are done. Otherwise, we can find a such that . Pick such of minimal order.

Claim 2: .
Proof: Consider . Since is of minimal order, . This means for some . Now we get So . Also, since , and so meaning since we are in a -group. This means , so let . Define . We firstly observe that since , and secondly we see that . So we conclude that and by the minimality of we have .

Now let . Since is of prime order, we know is cyclic. Furthermore, we know . To apply the induction hypothesis, we need to find an element of maximal order in . We can show that is such an element. By Cor. 1.3.13, for every where , . This means since if there is some in , then it generates all of , but . We now get that since , meaning . But the order is clearly at most , which is the next possible order, so , meaning indeed has maximal order.

Now we can apply the induction hypothesis, meaning there exists a subgroup such that . Now notice that there is a one-to-one correspondence between each and each . this is true since we can use the canonical projection and map and . We now use this to get meaning and .

Claim 3: .
Proof: Let . We then get by the previous paragraph. This in turn means that .

Finally, this means is injective, meaning

Every finite abelian group is the direct product of cyclic groups of prime-power order.

Proof:
We have which by Theorem means . Furthermore we know where and has maximal order. This means and is a -group, . By induction we get .

a) abelian and .
b) and abelian or .
c) , non-abelian, could be , but also the quaternion group. The quaternion group has 8 elements with relations Exercise: show that this is a group!
Subgroups: , , , , . We can show that these are all normal!

Let be abelian such that there is a subset such that , where . Then where is a finite abelian group.

Consider for some with factorization . Then there is a unique subgroup of order , namely .

Proof:
Clearly it's a subgroup of order . We know show uniqueness. Let . By Fermat, we know that for every meaning every element in is a multiple of , more concretely, . We also know so .