MAT301 Term Test 2 Solutions
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MAT301 Term Test 2 Solutions
Exercise 1. Let G be a group and suppose that H is a normal subgroup of G. Prove that the following statements are equivalent:
1. H is such that for every normal subgroup N of G satisfying H ≤ N ≤ G we must have N = G or N = H .
2. G/H has no non-trivial normal subgroups.
Proof. (⇒) Suppose that K ⊂ G/H is a non-trvial normal subgroup of G/H . Consider the canonical projection p : G ⇒ G/H . We know that the preimage p−1 (K) ⊂ G is a subgroup, it strictly contains H due to non-triviality in the quotient, and it is a normal subgroup: for every g ∈ G and k\ ∈ p−1 (K) we have
p(gk\g −1) = p(g)p(k\ )p(g)−1 ,
and p(k\ ) ∈ K, so p(gk\g −1) ∈ K as well, so the preimage is normal, which leads to a contradiction. (⇐) The idea is to consider the same projection. If N is a normal subgroup satisfying H ≤ N ≤ G, then
p(N) ⊂ G/H is a normal subgroup, so p(N) = {eH} or p(N) = G/H, which is equivalent to N = H or
N = G, respectively.
Exercise 2. Let G be the subgroup of GL4 (R) defined by
G = { (0(A) B(∗)) ,A,B ∈ GL2 (R) and ∗ ∈ Mat2×2 (R)} .
Let N be the subgroup of G defined by
N = { (0(I2)
I2(∗)) , ∗ ∈ Mat2×2 (R)} ,
where I2 ∈ GL2 (R) is the 2 × 2 identity matrix. Prove that N is a normal subgroup of G and give an isomorphism between G/N and a group that is not a quotient group.
Proof. Define ϕ : G → GL2 (R) × GL2 (R) by
ϕ ((0(A) B(∗))) = (A,B)
for all (0(A) B(∗)) ∈ G. The map ϕ is a homomorphism since for all ) , ) ∈ G, we have
)( ) = )
for some X3 ∈ Mat2×2 (R). Also, the homomorphism ϕ is surjective and its kernel is N . Therefore N is a normal subgroup of G and the map
G/N −→ GL2 (R) × GL2 (R)
gN -−→ ϕ(g)
is a well-defined isomorphism by the First Isomorphism Theorem.
Exercise 3. Let G be a group. Suppose that the quotient of G by one of its abelian normal subgroups is abelian. Prove that if H is a subgroup of G, then the quotient of H by one of its abelian normal subgroups is abelian. (Hint: Apply the Second Isomorphism Theorem.)
Proof. Let N be an abelian normal subgroup of G such that G/N is abelian. We will show that the subgroup H ∩ N of H is an abelian normal subgroup of H such that H/H ∩ N is abelian. Since H ∩ N ≤ N and N is abelian, we have that H ∩ N is abelian. By the Second Isomorphism theorem, we have that H ∩ N is a normal subgroup of H and
H/H ∩ N HN/N.
Finally, since HN/N ≤ G/N and G is abelian, we have that HN/N and therefore H/H ∩ N is abelian.
Exercise 4. Let p,q,r be distinct primes and let A be a finite abelian group of order pqr . Without using the classification of finite abelian groups, prove that A Z/pqrZ. (Hint: Show that A Z/pZ×Z/qZ×Z/rZ.)
Proof. By Cauchy’s Theorem for abelian groups there exist b,c,d ∈ A such that o(b) = p, o(c) = q, and o(d) = r . Let B = ⟨b⟩, C = ⟨c⟩, and D = ⟨d⟩ . It suffices to prove that A is the internal direct product A = B × C × D . Indeed, if this is the case then we have
A Z/pZ × Z/qZ × Z/rZ Z/pqrZ
since B Z/pZ, C Z/qZ, D Z/rZ, and p,q,r are mutually coprime. By the theorem on internal direct products, it suffices to show the following
1. B,C,D 司 A,
2. A = BCD, and
3. B ∩ C = {e} and BC ∩ D = {e} .
Item 1 is immediate since A is abelian. By Lagrange’s theorem, we have
|B ∩ C| I |B|, |C|
Since |B| = p and |C| = q are distinct primes, it follows that |B ∩ C| = 1. Therefore B ∩ C = {e} . Now, we
have
|BC| = = = pq .
Since A is abelian, we have BC = CB, and therefore BC is a subgroup of A. By Lagrange’s theorem, we have
|BC ∩ D| I |BC|, |D|
Since |BC| = pq and |D| = r are relatively prime, it follows that |BC ∩ D| = 1. Therefore BC ∩ D = {e} . Thus, Item 3 above holds. Finally, we have
|BCD| = |(BC)D| = = = pqr = |A|,
so BCD = A and Item 2 above holds.
Exercise 5. Let G be a finite group and let N be a normal subgroup of G such that gcd(|N|, |G/N|) = 1. Prove the following:
1. If H is a subgroup of G having the same order as G/N, then G = HN .
2. Let σ be an automorphism of G. Prove that σ(N) = N .
Proof. 1. Equivalently, we can prove that |HN| = |G| . But we know that |HN| = . As |H| =
|G/N| = |G|/|N|, we get
|HN| = |G||N| = |G|
The intersection H ∩ N is a subgroup in both H and N, so Lagrange implies that |H ∩ N| divides both |G/N| = |H| and |N|, which, together with the fact that both these orders are coprime, implies |H ∩ N| = 1, which finally yields |G| = |HN| .
2. As σ is an automorphism, σ(N) is a normal subgroup of G, let us denote it by N\ . Consider NN\ , observe that N ≤ NN\ , and gcd(|N|, |NN\ |/|N|) = 1, as (|NN\ |/|N|)|(|G|/|N|). Applying the above formula (or even the Second Isomorphism Theorem), we get
gcd(|N|, |NN\ |/|N|) = gcd(|N|, |N|/|N ∩ N\ |) = 1,
this can only happen if N ∩ N\ = N = N\ = σ(N).
2022-08-17