Algebra Qual Study Problems: 4

Final #4: (a) Find all Sylow subgroups of G, such that $|G| = 16$.
Proof-
Write $|G| = 16 = 2^4$. Then we can see there is only one Sylow 2 subgroup of order 16, which is the entire group

(b) $|G| = pq$ with p, q prime. Say that $q wlog. What is the cardinality of Sylow subgroups of G. Prove that any 2 distinct sylow subgroups intersect in the identity. Suppose in addition that all Sylow subgroups are normal, prove that G is isomorphic to the direct product of its Sylow Subgroups.
Proof- If G is a sylow-p subgroup then it must have order p, and a sylow-q subgroup must have order q. If P is a sylow p-subgroup and Q is a sylow q-subgroup, then $P \cap Q = \emptyset$ because an element cannot have order a power of p and order a power of q at the same time since p and q are distinct primes. Let $P_1, P_2$ be two sylow p-subgroups. Let $a\in P_1 \cap P_2$ then $\langle a\rangle \subset P_1 \cap P_2$. Since $|P_1| = |P_2| = p$ and they are p groups, the element a must also have order p. Thus $|\langle a\rangle | = p \Rightarrow P_1 = P_2$. Now assume that all Sylow subgroups are normal. Let P be a sylow p-subgroup and Q be a sylow q-subgroup. Then we have shown that $P \cap Q = \emptyset$ and $|\langle P\cup Q \rangle | = p\cdot q \Rightarrow \langle P\cup Q \rangle = G$, and since P and Q are normal in G we have by previous theorem that $G \cong P \times Q$.

Qual 95 #4, 08 #2 07 #4 (Hungerford 2.5 #9): If $|G| = p^nq$ with $p>q$ primes, then G contains a unique normal subgroup of index q.
Proof-
From Sylow’s First Theorem, G has a Sylow p-subgroup of order $p^n$. Let m be the number of Sylow p subgroups of G. By Sylows Third Theorem we have that $m | |G|$ and $m\equiv 1 \pmod{p}$. Then since $|G| = p^nq$ we have to have that either $m=1,q$ or $p^i$ for some $1\leq i \leq n$. If $m = q$ then $p|(q-1) \Rightarrow q-1 = pk$ for some $k\in \mathbb{Z}$. Then $pk+1 = q < p \Rightarrow k = 0 \Rightarrow q =1$ which is a contradiction. If instead we have that $m = p^i$ then $p | (p^i-1) \Rightarrow p = 1$, which is another contradiction. Therefore $m =1$ so that there is the only 1 Sylow p subgroup of G, call it H, which must be normal in G. Then by Lagrange, $|G| = [G:H]|H| \Rightarrow p^nq = [G:H] p^k \Rightarrow q = [G:H]$. Thus H has index q.

Qual 2005 #2, Final #5 (part 2 is Hungerford 2.5.11): 1) Exhibit 2 distinct Sylow 2 subgroups of $S_5$ and an element of $S_5$ conjugating one into the other.
Example –
Note that $|S_5|= 5! = 2^3 \cdot 3 \cdot 5$. Thus a sylow 2 subgroup has order 8. $\langle (24), (1234) \rangle \cong D_4$ and so is a Sylow 2 subgroup. Since any 2 Sylow subgroups are conjugate we can finish by working backwards. Conjugate the given subgroup by the element (12). So we have that $(12)(24)(12) = (14)$ and $(12)(1234)(12) = (1242)$ so that $\langle (14), (1342) \rangle$ is another distinct Sylow 2 subgroup with the desired property.

2) How many elements of order 7 are there in a simple group of order 168.
Proof-
Let $|G| = 168 = 2^3\cdot 3 \cdot 7$. By the Third Sylow Theorem, the number of Sylow 7-subgroups must divide $2^3\cdot 3$ and also must be congruent to 1 modulo 7. Thus the only possibilities are that the number of Sylow 7-subgroups is 1 or 8, but since G is simple we cannot have only one Sylow 7-subgroup otherwise it would be normal. Therefore there are 8 of them, each with 7 elements, with each non identity element having order 7 since a group with prime order must be cyclic. Thus there are $8(7-1)=48$ elements of order 7.

Final 2010 9 (Hungerford 2.5.10): Prove that a group of order 12 has a normal sylow subgroup.
Proof-
$|G| = 12 = 2^2\cdot 3$. By 3rd Sylow Theorem we know that there is either 1 Sylow 3 subgroup, or 4 Sylow 3 subgroups. If there is only one, then know it must be normal. Thus, suppose there are 4, $H_1,H_2,H_3,H_4$ each with order 3. Each of these subgroups has 3 elements, 1 being the identity. Thus the other 2 elements are of order 3. Together then we have $4(3-1) = 8$ elements of order 3 in G. Then G can only have one Sylow 2-subgroup which must be of order 4 because this gives us 12 total elements. Again, being the unique Sylow 2-subgroup gives that it is also normal.

Qual 2006 #4, 1999 #6: (1)Determine the units in $\mathbb{Z}[\sqrt{-5} ] = \{ a +b\sqrt{-5} : a,b \in \mathbb{Z}\}$.
Proof-
Define the map $N: \mathbb{Z}[\sqrt{-5} ] \rightarrow \mathbb{Z}$ by $a +b\sqrt{-5} \mapsto ( a +b\sqrt{-5} )( a -b\sqrt{-5}) = a^2 + 5b^2$. Note that $N(uv) = N(u)N(v)$ and claim u is a unit iff $N(u) = \pm 1$. Let u be a unit, then there exists $v$ such that $uv = 1 \Rightarrow 1 = N(1) = N(uv) = N(u)N(v) \in \mathbb{Z} \Rightarrow N(u) = \pm 1$. Then let $N(u) = 1$ where $u = a +b\sqrt{-5}$. Let $v = a -b\sqrt{-5} \Rightarrow N(u) = uv = a^2 + 5b^2 = 1 \Rightarrow$ u is a unit. The proof is similar if $N(u) = -1$ instead. Thus the claim is proved. Now we can prove (1). $a +b\sqrt{-5}$ is a unit iff $N(u) = (a +b\sqrt{-5} ) (a -b\sqrt{-5} ) = \pm 1 \Rightarrow a^2+5b^2 = 1 \Rightarrow b^2 = 0, a^2 = 1 \Rightarrow a = \pm 1$. Thus the only units are $\pm 1$ because the implications can be reversed. For (2), write $7 = (a +b\sqrt{-5}) (c +d\sqrt{-5} )$ Then $N(7) = N(a +b\sqrt{-5}) N(c +b\sqrt{-5}) \Rightarrow 49 = (a^2 + 5b^2) (c^2 + 5d^2)$. Wlog we have 2 possible outcomes: a) $a^2+ 5b^2 = 7$ and $c^2+5d^2 = 7$ or b)$a^2+5b^2 = 1$ and $c^2 + 5d^2 = 49$. The case of (a) gives a contradiction for any choice of b, while the (b) case gives a contradiction unless b = 0, in which case $a = \pm 1$. This gives that the only factorization of 7 is $7 = (7)(1)$ or $(7) = (-7)(-1)$ and so 7 is irreducible since 1 and -1 are units.

Qual 1999 #7: (a) Give an example of an integral domain R and ideals I and J of R such that $IJ \neq I \cap J$.
(b) Show that if $I + J = R$ then $IJ = I \cap J$.
Proof-
(a) Let $R = \mathbb{Z}, I = 2\mathbb{Z}, J = 4\mathbb{Z}$. Then $I \cap J = 4\mathbb{Z}$, but $IJ= 8\mathbb{Z}$. This is because in $\mathbb{Z}$, we have $\langle n \rangle + \langle m \rangle = \langle n,m \rangle = \langle gcd(n,m) \rangle$ but $\langle n \rangle \cap \langle m \rangle = \langle lcm(n,m) \rangle$.

(b) $IJ \subset I \cap J$ is always true because I and J are ideals.
Now let $x \in I \cap J$. Note that since we are given $I + J = R$ we can write $1_R = i + j$ for some $i\in I, j\in J$. Then $x = x\cdot 1_R = x(i + j) = xi + xj$. Then since x is in both I and J, $xi \in JI = IJ, xj \in IJ \Rightarrow x = xi + xj \in IJ$. This completes the proof.

Hungerford 3.1.15: (a) Give an example of a nonzero homomorphism $f: R \rightarrow S$ of rings with identity such that $f(1_R) \neq 1_S$.
(b) If $f:R \rightarrow S$ is an epimorphism then $f(1_r) = 1_S$.
(c) If $f:R \rightarrow S$ is a homomorphism of rings with identity and u is a unit in R such that $f(u)$ is a unit in S, then $f(1_R) = 1_S$ and $f(u^{-1}) = f(u)^{-1}$. Give an example to show that $f(u)$ need not be a unit in S even though u is a unit in R.
Proof-
a) Consider $f: \mathbb{Z}_3 \rightarrow \mathbb{Z}_6$ by $f(k) = 4k$. Then for $a,b\in \mathbb{Z}$ we have $f(a+b) = 4(a+b) = 4a +4b = f(a) + f(b)$ and $f(a)f(b) = 4a4b = 16ab = 4ab \pmod{6} = f(ab)$ but note that $f(1) = 4 \neq 1$ in $\mathbb{Z}_6$.

b) Let $s\in S$. Since f is surjective we have that $s = f(r)$ for some $r\in R$. Then $f(1_R) s = f(1_r) f (r) = f(r) = s$ and similarly $sf(1_R) = s$. Since the identity element is unique and $f(1_R)$ is acting as the identity we must have that $f(1_R) = 1_S$.

c) First note that $f(u) = f(1_R u ) = f(1_R) f(u)$. Then since $f(u)$ is given to be a unit we can multiply both sides by the inverse so that $1_s = f(1_R)$. Then using this fact we have $1_S = f(1_R) = f(uu^{-1}) = f(u) f(u^{-1})$ and similarly $1_S = f(u^{-1}) f(u)$. Then since inverses are unique and $f(u^{-1})$ is acting as the inverse of $f(u)$ we must have that $f(u^{-1}) = f(u)^{-1}$. Lastly, use the same map as in part a) for the counterexample and note that 2 is a unit $\mathbb{Z}_3$, but $f(2)=2$ is not a unit in $\mathbb{Z}_6$.

Qual 2013 #4: Find a maximal ideal in $\mathbb{C}[x]$ containing $f(x) = x^3-3x^2+2$. How many such ideals are there? In $\mathbb{R}[x]$ how many are there? Find $g(x) \in \mathbb{C}[x]$ such that $\langle f(x),g(x) \rangle \mathbb{C}[x]$.
Proof-
Recall some facts: If R is an integral domain and $0 \neq p(x),q(x) \in R[x]$ then 1) $deg(p(x)(q(x)) = deg(p(x)) + deg(q(x))$, 2) units of $R[x]$ are the units in R, 3) $R[x]$ is an ID also. Lastly, if F is a field then $F[x]$ is an ED which means it is a PID, UFD, and ID as well. Now, since $\mathbb{C}$ is a field we have that $\mathbb{C}[x]$ is a PID. Since any ideal is generated by an element, we know that $M =$ is maximal iff m is irreducible in $\mathbb{C}[x]$. Now we have $x^3-3x^2 + 2 = (x-1) (x-(1+\sqrt{3}))(x-(1-\sqrt{3}))$. Then we have $x^3-3x^2+2 \in \langle x-1 \rangle, \langle x-1-\sqrt{3} \rangle, \langle x-1+\sqrt{3} \rangle$. Since each factor is irreducible, each of these ideals is maximal. If $g(x) = 1$, then $\langle f(x),g(x) \rangle = \mathbb{C}[x]$.

Hungerford 3.4.12: A commutative ring with identity is local iff for all $r,x\in R$ we have $r+s =1_R \Rightarrow r \text{ or } s$ is a unit.
Proof-
$(\Leftarrow )$ Suppose that R is local and that $r+s= 1$. Suppose that r is not a unit. Since R is a local ring it has a unique max ideal, call it M. Then $r = 1-s \in M$ since all nonunits must be in M, and r is not a unit. If s was also a nonunit then we would have $(1-s)+s = 1 \in M$ which is a contradiction to the definition of maximal ideal. Therefore s must be a unit.
$(\Leftarrow )$ Let $1 = r + s \Rightarrow r \text{ or } s$ is a unit be true for all $r,s \in R$. Now, let r and s be nonunits. If $r + s$ is a unit then there exists a $x \in R$ such that $x(r+s) = 1 \Rightarrow x r + xs = 1$. Then by the assumption we have either $xr$ or $xs$ is a unit. This tells us that either r or s is a unit, which is a contradiction to assuming neither was a unit. Therefore, $r+s$ is a nonunit. If r is a nonunit then we know $xr$ is a nonunit for all x. Thus, we have that the sum of nonunits is a nonunit, and that nonunits absorb any element through multiplication. This gives that the set of all nonunits forms an ideal, which implies R is a local ring.

Qual 2013 #3: Prove that if R is a local ring and $\phi : R \rightarrow S$ is a surjective ring hom onto nonzero ring S, then S is local.
Proof-
Let R be a local ring and let M be the unique maximal ideal of R. Since the map is an epimorphism, ideals map to ideals and the inverse image of an ideal is an ideal which contains the kernel of the map. Now suppose that $\phi (M ) \subset N \subset S$ then $\phi^{-1}(\phi (M) \subset \phi^{-1}(N) \subset \phi^{-1}(S) =R$. This tells us that $M \subset \phi^{-1}(N) \subset R$. However we know M is maximal so that $\phi^{-1}(N) = M$ or R. Thus $N = \phi (M)$ or $S = \phi (R ) = N$ Therefore $\phi (M)$ is maximal. Next we show it is unique. If N is another maximal ideal in S, then we have $\phi^{-1}(N)$ is also max in R which by uniqueness of M gives that $\phi^{-1} (N) = M \Rightarrow N = \phi (M)$. Therefore S is a local ring.

Let $R = \mathbb{Z}$. Then $n\mathbb{Z}$ is maximal iff n is prime.
Proof-
$(\Rightarrow )$Let $p\mathbb{Z}$ be a maximal ideal and suppose $p\mathbb{Z} \subset J$. Then there exists an element which is in J but not in $p\mathbb{Z}$. Therefore $gcd(r,p) = 1$ so that $rx +py = 1$ for some $x,y \in \mathbb{Z}$. Then $py\in J$ and $rx\in J$ by definition of ideal. But this gives that $1\in J$ which means that $R = J$. Therefore $p\mathbb{Z}$ is maximal.
$(\Leftarrow )$ We know that the integers form a PID so if $m\mathbb{Z} \subset n\mathbb{Z}$ then $n |m$. So if $p\mathbb{Z} \subset n\mathbb{Z} \Rightarrow n|p \Rightarrow n =1 \text{ or } p$. Therefore $n\mathbb{Z} = \mathbb{Z}$ or $p\mathbb{Z}$ so that it is maximal.

Final 2013 #2: (a) Consider the ring $\mathbb{Z}$ of integers. Find a maximal ideal that contains the integer 45.
(b) Find a prime ideal in the integers that contains both 45 and 120
(c) Find an integer r such that $r\mathbb{Z} = 45\mathbb{Z} \cap 120\mathbb{Z}$.
Proof-
a) A maximal ideal in the integers is the same as a prime ideal. Since $45 = 3^2 \cdot 5$ we have that $45\in 3\mathbb{Z}$ which is a max ideal.

b) We can write $45 = 3^2\cdot 5$ and $120 = 2^3 \cdot 3\cdot 5$. Therefore both integers are contained in $3\mathbb{Z}$.

c) $45\mathbb{Z} \cap 120\mathbb{Z} = \{ x\in \mathbb{Z} | x = 45m = 120, \text{ where } n,m\in \mathbb{Z} \}$. But this is the same as the set $\{x\in \mathbb{Z} | 45|x \text{ and } 120 |x\}$. This set is generated by the lcm of 45 and 120. Therefore $45\mathbb{Z} \cap 120\mathbb{Z} = \langle lcm(45,120) \rangle = \langle 2^3\cdot 3^2 \cdot 5\rangle = 360\mathbb{Z}$.