## Algebra Qual Study Problems: 3

Example 1: How many non isomorphic abelian groups are there of order 1500?
Proof-
We have that $|G| = 1500 = 2^2 \cdot 3 \cdot 5^3$. Using this we can list all families of elementary divisors, which are : $\{2^2,3,5^3\},\{2,2,3,5^3\},\{2^2,3,5,5^2\},\{2^2,3,5,5,5\},\{2,2,3,5,5^2\},\{2,2,3,5,5,5\}$. This gives that there are exactly 6 non isomorphic abelian groups of order 1500 since each set corresponds to a unique (up to isomorphism) abelian group of order 1500.

Example 2: Find the invariant factors of $G = \mathbb{Z}_5 \bigoplus \mathbb{Z}_{15} \bigoplus \mathbb{Z}_{25} \bigoplus \mathbb{Z}_{36} \bigoplus \mathbb{Z}_{54}$.
Proof-
First we rewrite in terms of elementary divisors:
$G \cong \mathbb{Z}_5 \bigoplus ( \mathbb{Z}_{5} \bigoplus \mathbb{Z}_{3} ) \bigoplus \mathbb{Z}_{5^2} \bigoplus ( \mathbb{Z}_{2^2} \bigoplus \mathbb{Z}_{3^2} )\bigoplus ( \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3^3} )$. Thus we have the elementary divisors being $\{2,2^2,3,3^2,3^3,5,5,5^2\}$. Ordering them in the correct matrix formation we get:
$2^0,3,5$
$2^1,3^2,5$
$2^2,3^3,5^2$
Multiplying across gives us the invariant factors, which are $15,90,2700$.

Hungerford 2.2.12:(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:
$\mathbb{Z}_{32} \bigoplus \mathbb{Z}_{3}, \mathbb{Z}_{2} \mathbb{Z}_{3} \bigoplus \mathbb{Z}_{16}, \mathbb{Z}_{3} \bigoplus \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{8}, \mathbb{Z}_{8} \bigoplus \mathbb{Z}_{2}$, $\mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3}, \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3}$, $\mathbb{Z}_{4} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3}, \mathbb{Z}_{2} \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3}$

Hungerford 2.2.9: 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. Then $(np) \cdot p = np^2 = 0$. If there was some $m 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 $p^2$.

Midterm #5: G an abelian group of order 252. Find all possible primes such that G can contain a subgroup isomorphic to \mathbb{Z}_p \bigoplus \mathbb{Z}_p$for some prime p. Give an example where it doesn’t. Proof- By the classification theorem we can find all such groups G by listing out possible elementary divisors. $|G| = 252 = 2^2 \cdot 3^2 \cdot 7$ which gives the possible elementary divisors to be: $\{2^2,3^2,7\},\{2,2,3^2,7\},\{2^2,3,3,7\},\{2,2,3,3,7\}$. Writing them out in the correct way we can get invariant factors. $\{2^2,3^2,7\} \rightarrow \{252\}$ $\{2,2,3^2,7\} \rightarrow \{2,126\}$ $\{2^2,3,3,7\} \rightarrow \{3,84\}$ $\{2,2,3,3,7\} \rightarrow \{6,42\}$ From this we can see the only possible primes are 2 and 3. As a counterexample, $\mathbb{Z}_{252}$ is cyclic and thus every subgroup is cyclic. Therefore it cannot contain a subgroup isomorphic to $\mathbb{Z}_p \bigoplus \mathbb{Z}_p$ because that group is not cyclic. Hungerford 2.2.1: Show that a finite abelian group that is not cyclic contains a subgroup which is isomorphic to $\mathbb{Z} \bigoplus \mathbb{Z}$ for some prime p. Proof- Let G be a finite abelian group. Then we have$G \cong \mathbb{Z}_{p_1^{k_1}} \bigoplus \mathbb{Z}_{p_2^{k_2}} \bigoplus \cdots \bigoplus \mathbb{Z}_{p_r^{k_r}}$ 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 $G \cong \mathbb{Z}_{p_1^{k_1}} \bigoplus \mathbb{Z}_{p_2^{k_2}} \bigoplus \cdots \bigoplus \mathbb{Z}_{p_r^{k_r}}$. Theorem 2.4.3: If a group G acts on a set S then $| \bar{x} | = [G:G_x]$, where $\bar{x}$ is the orbit of x, and $G_x$ is the stabilizer subgroup of x. Proof- We can exhibit a bijection between the two sets which gives the cardinality to be the same. We have that, $gG_x = hG_x \Leftrightarrow h^{-1}gG_x = G_x \Leftrightarrow h^{-1}g \in G_x \Leftrightarrow h^{-1}gx = x \Leftrightarrow gx = hx$. This tells us that the association: $gG_x \rightarrow gx$ is injective, and it is obviously surjective, so that the cardinalities are equal. Hungerford 2.4.3: If a group G contains an element a having exactly 2 conjugates, then G has a proper normal nontrivial subgroup N. Proof- Let G be a group and let $a\in G$ have exactly 2 conjugates. That is, there are 2 elements in the conjugacy class (orbit) $\bar{a} = \{gag^{-1} | g\in G\}$ of a. By theorem 4.3, $|\bar{a} | = 2 = | \{gag^{-1} | g\in G\}| = [G:G_a]$, but by previous theorem this gives that $G_a$ is normal in G. If $G_a$ were to be trivial then that would mean that G has 2 elements so is the same as $\mathbb{Z}_2$ and so is abelian. But in an abelian group, every conjugacy class has 1 element so this is a contradiction. Thus $G_a = N$ is nontrivial and normal in G. October 2004 Qual #5 (Corollary 2.4.9): Let G be a finite simple group having a subgroup of index n. Show that G is isomorphic to a subgroup of $S_n$. Proof- Let H be a subgroup of G with index n and define the group action of G on the set of left cosets: $G \times G/H \rightarrow G/H$ by $(g_1,g_2H)\mapsto (g_1g_2)H$. This is clearly a group action. Then we have the induced homomorphism: $\phi G \rightarrow A(G/H)$ given by $g\mapsto \tau_g$, where $\tau_g :G/H \rightarrow G/H$ by $\tau_g (xH) = gxH$. Now suppose that $g\in Ker(\phi ).$ Then we have $\tau_g = id_{G/H} \Rightarrow \tau_g (xH) = xH = gxH \forall x\in G$. In particular we have that it is true for $x= e$, so that $geH = eH = H \Rightarrow g \in H$. This tells us that $Ker(\phi ) \leq H$. Note t hat G is simple, so that the Kernel must be trivial since it is a subset of H. Now by the first isomorphism theorem we have $G/Ker(\phi ) \cong Im( \phi ) \leq A(G/H)$. Note that since $|G/H| = n$ we have that $A(G/H) \cong S_n$. Thus $G/Ker(\phi ) \cong G/\{e\} \cong G \cong Im(\phi ) \leq A(G/H) \cong S_n$. April 2004 #2: Let G be a group. If Aut(G) is cyclic then G is abelian. Proof- First recall that the inner automorphisms of G, denoted $Inn(G) = \{\tau_g :G \rightarrow G | \tau_g (h ) = ghg^{-1}\}$ and that $Inn(G) \leq Aut(G)$ so that $Inn(G)$ is also cyclic. If G acts on G by conjugation, then the induced homomorphism $\phi : G \rightarrow Aut(G)$ has $Ker(\phi ) = C(G)$ and $Im(\phi ) = Inn(G)$ so that $G/C(G) \cong Inn(G)$. This gives that $G/C(G)$ is cyclic which tells us that G is abelian. 2011 Qual #5: Suppose that H acts on a set S. Given $s\in S$ define the H-orbit of s. Prove that if $s'\in S$ is another element then either $s'$ is in the orbit of s or that the orbits of s and s’ are disjoint. Proof- Suppose that $s'$ is not in the orbit of s, and suppose there is some $x\in \bar{s'} \cap \bar{s}$. Then we can write $x= h_1s'=h_2 \Rightarrow s' = h_1^{-1}h_2s$. This tells us that $s'$ is in the orbit of s which is a contradiction. 2013 Qual #2(a) (Hungerford 2.4.6): Consider $S_n$ acting on the set $S = \{1,2,\ldots , n\}$ by permutations. What is the orbit of i? What is the stabilizer of i? Prove that the stabilizer subgroup of any two elements of S are conjugate. Proof- Orbit of i: $\bar{i} = \{ \psi i | \psi \in S_n\} = S$ Stabilizer of i: Claim that $G_i \cong S_{n-1}$. First note that $[S_n : G_i] = |\bar{i} | = n \Rightarrow |G_i| = (n-1)!$, which comes from the fact that $|S_n| = [S_n : G_i] |G_i|$. Consider $S_{n-1}$ as the set of permutations on $S\backslash \{i\}$. Then clearly these permutations fix i and are a subgroup isomorphic to $S_{n-1}$. Next, we show that $G_i$ is conjugate to $G_j$. We know that for any i,j that exists $(ij) \in S_n$ such that $(ij) i = j$ \Rightarrow (ji) j = i$, since the inverse of (ij) is (ji). Claim that $(ji)G_j(ij) = G_i$. Let $\tau \in G_i$. Then $\tau i = i$. Note that $\tau = (ji) ( (ij) \tau (ji) ) (ij)$, while $(ij) \tau (ji) \in G_j$ because $(ij) \tau (ji) j = (ij) \tau i = (ij) i = j$. This gives that $\tau \in (ji)G_j(ij)$. Thus $G_i \subset (ji)G_j(ij)$. Now let $\tau \in (ji)G_j(ij)$. Then we can write $\tau (ji) \psi (ij)$ for some $\psi \in G_j$. Then $\tau i = (ji) \psi (ij) i = (ji ) \psi j = (ji) j = i$. This tells us that $\tau \in G_i$. Thus we have subset inclusion so that the two sets are conjugate.