Algebra Solutions: Hungerford II.2

1) Show that a finite abelian group that is not cyclic contains a subgroup which is isomorphic to \mathbb{Z}_p \bigoplus \mathbb{Z}_p for some prime p.
Proof-
Let G be a finite abelian group. Then we haveG \cong \mathbb{Z}_{p_1^{k_1}} \bigoplus \mathbb{Z}_{p_2^{k_2}} \bigoplus \cdots  \bigoplus \mathbb{Z}_{p_r^{k_r}} where p_i is prime and k_i is a positive integer. If every prime is distinct, then all the powers of the primes are relatively prime so that G \cong \mathbb{Z}_{p_1^{k_2}p_1^{k_2}\cdots p_r^{k_r}}, which is cyclic. So instead we must have some case of p_i = p_j. WLOG suppose that p_1 = p_2 and call them both p. Then consider that \mathbb{Z}_{p^{k_i}} = \langle 1 \rangle where i = 1,2. Note that p^{k_i-1} \in \mathbb{Z}_{p^{k_i}} so that \langle p^{k_i-1} \rangle is a cyclic subgroup of \mathbb{Z}_{p^{k_i}}. Since pp^{k_i-1} = 0 in \mathbb{Z}_{p^{k_i}} the order of p^{k_i-1} must divide p. This is only possible if | p^{k_i-1}| = p which means \langle  p^{k_i-1} \rangle is a cyclic subgroup of \mathbb{Z}_{p^{k_i}} of order p, and thus isomorphic to \mathbb{Z}_p. Then we have that \langle p^{k_1-1} \rangle \bigoplus \langle p^{k_2-1} \rangle \bigoplus \{0\} \bigoplus \cdots \bigoplus \{0\} is a cyclic subgroup of G \cong \mathbb{Z}_{p_1^{k_1}} \bigoplus \mathbb{Z}_{p_2^{k_2}} \bigoplus \cdots  \bigoplus \mathbb{Z}_{p_r^{k_r}}.

9) How many subgroups of order p^2 does the abelian group \mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2} have?
Proof-
By the classification theorem we know that if |G| = p^2 then G \cong \mathbb{Z}_p \bigoplus \mathbb{Z}_p or \mathbb{Z}_{p^2}. So we first need to know how many unique subgroups of \mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2} there are which are isomorphic to \mathbb{Z}_p \bigoplus \mathbb{Z}_p. Consider the elements in \mathbb{Z}_{p^2}
1,2,3 \ldots ,p ,p+1, ,p+2,p+3, \ldots ,2p, 2p+1, \ldots ,(p-1)p , (p-1)p+1, (p-1)p+2, \ldots p \cdot p. Now we need to know which of these has order p. All elements that are multiples of p other than the identity, have order p. Consider an element np, where 1\leq n<p. Then (np) \cdot p = np^2 = 0. If there was some m <n such that npm = 0 we would have that npm = kp^2 \Rightarrow nm = kp \Rightarrow p | nm which is a contradiction because gcd(p,n)=gcd(p,m) = 1. There are p-1 multiples of p and so there are p-1 elements of order p in \mathbb{Z}_{p^2}.
Next we find the number of elements of order p in \mathbb{Z}_{p^3}. In a similar manner we get that all multiples of p^2 have order p and that there are p-1 such elements in \mathbb{Z}_{p^3}.
Lastly we need how many elements there are of the form (x,y) \in \mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2} which have order p. Note that |(x,y)| = p \text{ iff } lcm(|x|,|y|) = p. This can only happen in a few cases. If either order is 1, then the other must be p, or they can both be p. If |x|= p, then x can be p-1 elements. In order for the least common multiple to be p we need that |y| = 1 or |y| = p-1. So each of the p-1 elements that x can be can be paired with the identity of y, giving p-1 choices. Then each of the p-1 elements that x can be can be paired with the p-1 choices of y, giving (p-1)(p-1) choices. Lastly if x is the identity it can be paired with p-1 more choices from y. This gives a total of p-1+(p-1)(p-1) + p-1 = (p-1)( 1 + p-1 + 1) = (p-1)(p+1) = p^2-1 elements of order p in \mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2}. Note that \mathbb{Z}_p \bigoplus \mathbb{Z}_p has order p^2 and has p^2-1 elements of order p so that there can only be one subgroup of \mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2}. There are not enough elements of order p to make up another subgroup.
Next we need to know how many unique subgroups there are of \mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2} which are isomorphic to \mathbb{Z}_{p^2}. To do this we need a lemma: If there are n elements of order d in a group G, then there are \frac{n}{\phi (d)} cyclic subgroups of order d, where \phi is the Euler Phi function, which counts the number of positive integers relatively prime and less than the input number. Thus by the lemma it is enough to count the number of elements of order p^2 in \mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2}. Again an element (x,y) has order p^2 iff lcm(|x|,|y|) = p^2. This can only happen if either |x| = p^2 or |y| = p^2. After an annoying counting argument we get that there are (p^2-p)(p^2+p) such elements so that by the lemma we have \frac{(p^2-p)(p^2+p)}{\phi (p^2)} = p^2 +p subgroups of type \mathbb{Z}_{p^2}. Thus in total there are p^2 + p + 1 subgroups of order p^2.

12)(a) What are the elementary divisors of the group \mathbb{Z}_2 \bigoplus \mathbb{Z}_9 \bigoplus \mathbb{Z}_{35}. Do the same for \mathbb{Z}_{26} \bigoplus \mathbb{Z}_{42} \bigoplus \mathbb{Z}_{49}  \bigoplus \mathbb{Z}_{200} \bigoplus \mathbb{Z}_{1000}.
Proof-
For the group \mathbb{Z}_2 \bigoplus \mathbb{Z}_9 \bigoplus \mathbb{Z}_{35} we have that 2 = 2, 9 = 3^2, 35 = 5 \cdot 7. Thus the elementary divisors are 2, 3^2,5,7. To get the invariant factor we need only multiply the elementary divisors together because they are all relatively prime. Therefore the invariant factor is 2 \cdot 3^2 \cdot 5 \cdot 7 = 630. Then we can say finally that the group \mathbb{Z}_2 \bigoplus \mathbb{Z}_9 \bigoplus \mathbb{Z}_{35} \cong \mathbb{Z}_2 \bigoplus \mathbb{Z}_{3^2} \bigoplus \mathbb{Z}_{5} \bigoplus \mathbb{Z}_{7} \cong \mathbb{Z}_{630}.
In a similar manner we have the elementary divisors for \mathbb{Z}_{26} \bigoplus \mathbb{Z}_{42} \bigoplus \mathbb{Z}_{49}  \bigoplus \mathbb{Z}_{200} \bigoplus \mathbb{Z}_{1000} are 2,2,2^3,2^3,3,5^2,5^3,7,7^2,13. Because we have 5 unique primes and 2 appearing the most amount of times, 4, we arrange the elementary factors into 4 rows and 5 columns ordering from least to greatest filling in with 0 powers where necessary:
2^1 \cdot 3^0 \cdot 5^0 \cdot 7^0 \cdot 13^0
2^1 \cdot 3^0 \cdot 5^0 \cdot 7^0 \cdot 13^0
2^3 \cdot 3^0 \cdot 5^2 \cdot 7^1 \cdot 13^0
2^3 \cdot 3^1 \cdot 5^3 \cdot 7^2 \cdot 13^1
Lastly, multiply across to get the invariant factors of 2,2,1400,1911000. Thus we finally have:
\mathbb{Z}_{26} \bigoplus \mathbb{Z}_{42} \bigoplus \mathbb{Z}_{49}  \bigoplus \mathbb{Z}_{200} \bigoplus \mathbb{Z}_{1000} \cong \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{13} \bigoplus \mathbb{Z}_{2}  \bigoplus \mathbb{Z}_{3} \bigoplus \mathbb{Z}_{7} \bigoplus \mathbb{Z}_{7^2} \bigoplus \mathbb{Z}_{2^3} \bigoplus \mathbb{Z}_{5^2} \bigoplus \mathbb{Z}_{5^3} \cong \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{1400} \bigoplus \mathbb{Z}_{1911000}.

(b) Determine up to isomorphism all abelian groups of order 64 and 96.
Proof- 64 = 2^6 so we have the following abelian groups. \mathbb{Z}_{64}, \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{32}, \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{16}, \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{16}, \mathbb{Z}_{8} \bigoplus \mathbb{Z}_{8}, \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{8}, \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{8}, \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{4}, \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{4}, \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{4}, \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2}.
Next we have that 96 = 2^5 \cdot 3 so that any abelian group of that order is isomorphic to one of the following:
\mathbb{Z}_{32} \bigoplus \mathbb{Z}_{3}, \mathbb{Z}_{2} \mathbb{Z}_{3} \bigoplus \mathbb{Z}_{16}, \mathbb{Z}_{3} \bigoplus \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{8}, \mathbb{Z}_{8} \bigoplus \mathbb{Z}_{2}, \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3}, \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3}, \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3}, \mathbb{Z}_{2} \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3}

13) Show that the invariant factors of \mathbb{Z}_m \bigoplus \mathbb{Z}_n are gcd(m,n) and lcm(m,n) if gcd(m,n) >1 and mn if gcd(m,n) = 1.
Proof-
First assume that gcd(m,n) = d >1 . We can factor m and n so that m = p_1^{k_1}\cdots p_r^{k_r} and n = p_1^{l_1}\cdots p_r^{l_r} where p_i is prime and k_i,l_i \geq 0. Then lcm(m,n) = p_1^{a_1}\cdots p_r^{a_r} and gcd(m,n) = p_1^{b_1}\cdots p_r^{b_r} where a_i = max\{k_i,l_i\} and b_i = min\{k_i,l_i\}. Then we have:
\mathbb{Z}_m \bigoplus \mathbb{Z}_n \cong \mathbb{Z}_{p_1^{k_1}} \bigoplus \cdots \bigoplus \mathbb{Z}_{p_r^{k_r}} \bigoplus \mathbb{Z}_{p_1^{l_1}} \bigoplus \cdots \bigoplus \mathbb{Z}_{p_r^{l_r}}.
Now we have it written as elementary divisors. To get the invariant factors we follow the algorithm and arrange the primes as follows: 2 rows because at most a prime can appear twice, r rows for r primes.
min(p_1^{k_1},p_1^{l_1}) min(p_2^{k_2},p_2^{l_2}) ... min(p_r^{k_r},p_r^{l_r})
max(p_1^{k_1},p_1^{l_1}) max(p_2^{k_2},p_2^{l_2}) ... max(p_r^{k_r},p_r^{l_r})
Then we multiply across as before. However here we note that the multiplication is just:
p_1^{b_1} \cdots p_r^{b_r}
p_1^{a_1} \cdots p_r^{a_r}
So these are the invariant factors. Thus we have:
\mathbb{Z}_m \bigoplus \mathbb{Z}_n \cong \mathbb{Z}_{p_1^{k_1}} \bigoplus \cdots \bigoplus \mathbb{Z}_{p_r^{k_r}} \bigoplus \mathbb{Z}_{p_1^{l_1}} \bigoplus \cdots \bigoplus \mathbb{Z}_{p_r^{l_r}} \cong \mathbb{Z}_{p_1^{b_1} \cdots p_r^{b_r}} \bigoplus \mathbb{Z}_{p_1^{a_1} \cdots p_r^{a_r}} \cong \mathbb{Z}_{gcd(m,n)} \bigoplus \mathbb{Z}_{lcm(m,n)}. Note that the gcd always divides the lcm so that this is written in terms of invariant factors. If instead the gcd(m,n) = 1 then we have \mathbb{Z}_m \bigoplus \mathbb{Z}_n \cong \mathbb{Z}_{mn} where lcm(m,n) = mn so the invariant factor of \mathbb{Z}_m \bigoplus \mathbb{Z}_n is mn.

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