Tutorial 4A

1 Intersection of subgroup is a subgroup.

Clear that \(\bigcap_{i\in I} H_i \subseteq G\). To show it’s a subgroup, show closure and inverse is inside.

Take any \(a,b \in \bigcap_{i\in I} H_i\), expanding using logic we have \(a,b\in H_i\) for all \(i\in I\), then since each \(H_i\) is a subgroup \(ab \in H_i\) for all \(i\in I\) so \(ab \in \bigcap_{i\in I} H_i\). Closure shown. Similar argument shows inverse is inside the intersection.

2 Show these are equivalent

\(a^{-1}b\in H\),

\(b \in aH\),

\(aH = bH\).

Suppose (i), then exists \(h \in H\) such that \(a^{-1}b = h\), then \(b = ah\) which entails that \(b \in aH\). So (i) implies (ii).

Suppose (ii), then exists \(h\in H\) such that \(b = ah\). Take a \(ah'\in aH\), then \(ah' = bh^{-1}h' \in bH\). So \(aH \subseteq bH\). Similarly take \(bh'' \in bH\), then \(bh'' = ahh'' \in aH\). So \(aH = bH\) and we have (ii) implies (iii).

Suppose (iii), then take any \(x \in aH = bH\), exists \(h, h'\) such that \(x = ah = bh'\), then \[\begin{align*} a^{-1} x = h &= a^{-1}bh' \\ h{h'}^{-1} &= a^{-1}b \end{align*}\] then we have (i).

3 Equivalence relations

\(H < G\), \(\forall a,b \in G.~ a \equiv b \pmod{H} \iff a^{-1}b \in H\).

We have \(a \equiv b \pmod{H} \iff a^{-1}b \in H \iff aH = bH\). Relation is logically equivalent to a strict equality of cosets so it’s clear.

3.1 Show equivalence class.

Fix \(a\in G\).

Let \(b\) such that \(a\equiv b \pmod{H}\), that is \(a^{-1}b \in H\). By previous question \(b\in aH\), so \([a] \subseteq aH\). Let \(b\in aH\) by previous question \(a^{-1}b \in H\).

4 Well-defined

Show that \(f : G/H \to H \setminus G\) given by \(f(gH) = Hg^{-1}\) is a well-defined bijective function.

Using a symmetric argument as question 2, I can show the logical equivalence that \(ab^{-1} \in H \iff a \in Hb \iff Ha = Hb\). (But I’m lazy to).

Notice that the definition of \(f\) could possibly ambiguously match multiple candidates for \(g\) in the expression \(f(gH) = Hg^{-1}\). However, we show that if \(g, g'\in G\) such that \(g\ne g'\) and \(gH = g'H\), then \(f(gH) = Hg^{-1} = Hg'^{-1} = f(g'H)\).

Now \(gH = g'H \implies g^{-1}g' \in H \implies Hg^{-1} = Hg'^{-1}\).

Then produce an inverse function, shouldn’t be too hard.

again

Consider the diagram \[\require{AMScd} \begin{CD} G @>{(\cdot)^{-1}}>> G \\ @VV{q}V @VV{q}V \\ G/H @>f>> H\setminus G \end{CD} \] \(f\) is defined such that this diagram commutes, so do a diagram chase.

5 Chain rule

Let \(K < H < G\). Show that \([G:K]=[G:H] [H:K]\).

A partition counting argument is very hard to formalise.

6 More Linear Algebra

Let \(G\) be a finite subgroup of \(\operatorname{GL}(m, \mathbb{C})\) of order \(n\). Let \(g\in G\).

Show that \(g^n\) is identity matrix.

Generate a cyclic subgroup based on the powers of \(g\), let it have order \(k\), then \(g^k = I\) and by Lagrange theorem, \(k\mid n\), so \(g^n = I\).
Show that \(g\) is diagonalisable and every eigenvalue \(\lambda\) is a \(n\)-th root of unity.

By part i the polynomial \(f(T) = T^n - 1 \in \mathbb{C}[T]\) kills \(g\). Therefore the minimal polynomial \(\mu \mid f\). Notice that in \(\mathbb{C}\) \(f(T)\) splits fully into linear factors, so \(\mu\) splits into linear factors, this property characterises diagonalisable operators so \(g\) is diagonalisable. We also know that any polynomial that kills \(g\) kills its eigenvalues, so \(f(\lambda) = 0\) for any eigenvalue \(\lambda\).

7 \(K = n\mathbb{Z}\) is a subgroup of \(Z\).

Subgroup \(H\) is an ideal so it’s \(d\mathbb{Z}\) for some \(d\in \mathbb{Z}\). If \(d\mid n\), then \(dk = n \implies \underbrace{d + d + \dots + d}_{k\text{ times}} = n\) so \(n \in H\) so \(n\mathbb{Z}\subseteq H\).
Generate the subgroup with \(d\).
Skipping

Tutorial 4B

8 Free groups

Refer to hint resource for construction. Key concepts (reducibility , reducibility in one step) I already understand.

9 UMP of free groups

Draw the damn diagram.

10 Orders of subgroups

We have \(H < G\) and \(K < G\) and \(\gcd(\left\lvert H \right\rvert,\left\lvert K \right\rvert) = 1\). Well \(H \cap K\) is a subgroup too and its order divides both of that of \(H\) and \(K\) which entails that it has order 1.

11 Orders of subgroups

\(p\mid n\) so \(p \leq n\).