MT 1 Review (Abstract Algebra)

Review this vocab:

Group
Abelian Group
Group Morphism: AKA Group Homomorphism (or Group Morphism)
Isomorphism
Subgroup
Kernel
Group Action
Stabilizers
Orbit
Faithful Action
Centralizer
Normalizer
Cyclic Group
#todo/school : Do the vocab words above and review 📅 2025-01-29 ✅ 2025-01-29
#todo/school : Do the practice problems (1-6) in 1 hour and see how you did. See this link 📅 2025-01-30 ✅ 2025-01-29

MT Practice

# 1-3 are practice computations. #4-9 are practice proofs.

Overall here's how I did:

I edited the order formula for 1a but overall got the answer right
1b is mostly right (small numeric errors) but the cyclic group part was wrong.
2 try to make sure you know the cycle decomposition algorithm (pg 44 on the textbook)
3 I DNF'ed
But the proofs looked perfect!

1

Question

a. Compute the Order of each element in $\frac{Z}{10 Z}$
b. Compute the order of each element in ${(\frac{Z}{10 Z})}^{\times}$ . Is this a Cyclic Group? If so, give a generator.
c. Draw the lattice of subgroups of the additive group $\frac{Z}{10 Z}$

Proof

a. For each of these the order is just $| \overset{―}{a} | = \frac{10}{gcd (a, 10)}$

$| \overset{―}{0} | = 1$
$| \overset{―}{1} | = 10$
$| \overset{―}{2} | = 5$
$| \overset{―}{3} | = 10$
$| \overset{―}{4} | = 5$
$| \overset{―}{5} | = 2$
$| \overset{―}{6} | = 5$
$| \overset{―}{7} | = 10$
$| \overset{―}{8} | = 5$
$| \overset{―}{9} | = 10$

$| \overset{―}{0} | = \infty$
$| \overset{―}{1} | = 1$
$| \overset{―}{2} | = \infty$
$| \overset{―}{3} | = 4$
$| \overset{―}{4} | = \infty$
$| \overset{―}{5} | = \infty$
$| \overset{―}{6} | = \infty$
$| \overset{―}{7} | = 4$
$| \overset{―}{8} | = \infty$
$| \overset{―}{9} | = 2$
It is cyclic since ${(\frac{Z}{10 Z})}^{\times} = ⟨ \overset{―}{9} ⟩$ .

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2

Question

Let $σ \in S_{6}$ be the permutation $σ = (1 4 5 6) (2 1 3 5)$ . Write $σ$ as a product of disjoint cycles and compute its order.

Proof

To compute, let's do the permutations from right to left:

$σ$	1	2	3	4	5	6
1	3	1	5	4	2	6
2	3	4	6	5	2	1
Thus $σ \equiv (3 4 6 5 2 1) \equiv (1 3 6) (2 4 5)$ . Thus, then $	\sigma	= \text{lcm}(3,3) = 3$.

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3

Question

Let $A \leq S_{3}$ be the Subgroup generated by the transposition $(1 2)$ .
a. Determine the Centralizer of $A$ in $S_{3}$ .
b. Determine the Normalizer of $A$ in $S_{3}$ .
c. Determine the Center of $A$ in $S_{3}$ .

Proof

a. We want to get ${σ \in S_{3} | σ (1 2) = (1 2) σ \land σ 1 = 1 σ}$ . Namely, we can commute permutations only when they don't share the same number. Using our table, it's the set ${(), (1 2), (1 3 2)}$ .

b. We want to get ${σ \in S_{3} | σ A σ^{- 1} = A}$ , which in this case is all ${σ \in S_{3} | σ 1 σ^{- 1} \in A, σ (1 2) σ^{- 1} \in A}$ , which happens only when we check each element:

$g \in S_{3}$	Result w/ $x = (1 2)$
$e$	$e (1 2) e^{- 1} = (1 2)$
$(1 2)$	$(1 2) (1 2) (1 2)^{- 1} = (1 2)$
$(1 3)$	$(1 3) (1 2) (1 3)^{- 1} = (1 3) (1 2) (3 1) = (2 3)$
$(2 3)$	$(2 3) (1 2) (2 3)^{- 1} = (2 3) (1 2) (3 2) = (1 3)$
$(1 2 3)$	$(1 2 3) (1 2) (1 2 3)^{- 1} = (1 2 3) (1 2) (1 3 2) = (2 3)$
$(1 3 2)$	$(1 3 2) (1 2) (1 3 2)^{- 1} = (1 3 2) (1 2) (2 3 1) = (1 2)$
Thus $N_{G} (A) = {(), (1 2), (1 3 2)}$ .

c. We want to find $Z (A) = C_{A} (A) = {σ \in A | σ x = x σ \forall x \in A}$ . Using our table then $Z (A) = {(), (1 2)}$ in this case since there's only so many elements that can't commute.

(in case on the exam, look at the table method for finding these -izers)

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4

Question

Prove that every Group $G$ contains a unique identity element.

Proof

Assume for contradiction that it didn't have a unique identity. Label them $e_{1}, e_{2} \in G$ . Now since $e_{1}$ is the identity then we can left-multiply it with $e_{2}$ and get back $e_{2}$ :

e_{1} ⋆ e_{2} = e_{2}

Similarly, since $e_{2}$ is the identity then we can right-multiply it with $e_{1}$ to get:

e_{1} ⋆ e_{2} = e_{1}

Thus $e_{1} = e_{2}$ is a contradiction. The identity must be unique!

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5

Question

Suppose $ϕ : G \to H$ is a Homomorphism (or Group Morphism). Prove that $ϕ (1_{G}) = 1_{H}$ .

Proof

Let $g \in G$ be arbitrary:

\begin{aligned} ϕ (g) & = ϕ (g 1_{G}) \\ = ϕ (g) ϕ (1_{G}) \end{aligned}

Now we can do left-cancellation to get that:

1_{H} = ϕ (1_{G})

as desired.

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6

Question

Suppose a group $G$ has no nontrivial proper Subgroup. Prove $G$ is cyclic.

Proof

Suppose $G$ has no nontrivial proper subgroup; thus only the trivial (proper) subgroup exists for $G$ .

Assume for contradiction that $G$ is not cyclic. Then $G$ has at least two generators $t_{1}, t_{2}$ where $G = ⟨ t_{1}, t_{2}, \dots ⟩$ . Now consider the subgroups $⟨ t_{1} ⟩$ and $⟨ t_{2} ⟩$ . Notice that if we had more generators then $G$ has more than one proper subgroup, such as $⟨ t_{1} ⟩$ , creating a contradiction. Thus, we can't have 3 or more generators.

But what about just $G = ⟨ t_{1}, t_{2} ⟩$ ? Even then our subgroups $⟨ t_{1} ⟩$ and $⟨ t_{2} ⟩$ are proper subgroups that aren't trivial, creating another contradiction. Thus then $G = ⟨ t ⟩$ for some generator $t$ , and thus $G$ is cyclic as it's composed of only one generator.

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7

Question

Suppose $G$ is an Abelian Group and $H, K \leq G$ are Subgroups. Prove $H \cap K$ is a subgroup of $G$ . Give an explicit example in which $H \cup K$ is not a subgroup of $G$ .

Proof

To show $H \cap K$ is a subgroup, first notice that $H \cap K \subseteq G$ since $H \subseteq G$ and $K \subseteq G$ .

To show closure, let $a, b \in H \cap K$ . Then $a, b \in H$ and $a, b \in K$ . By these $H, K$ being subgroups themselves, then $a b \in H$ and $a b \in K$ . Thus $a b \in H \cap K$ as desired.

To show the inverse is in the set, let $a \in H \cap K$ , so then $a \in H$ and $a \in K$ . By $H, K$ being subgroups then $a^{- 1} \in H$ and $a^{- 1} \in K$ , so then $a^{- 1} \in H \cap K$ as desired.

(Note that you don't actually need $G$ to be an Abelian Group in this case)

To show that $H \cup K$ may not be a subgroup of $G$ , consider $G = D_{8}$ and $H = ⟨ r ⟩$ and $K = ⟨ s ⟩$ . Notice $H \cup K = {1, r, r^{2}, r^{3}, s}$ which is not a subgroup of $G = D_{8}$ as it is not closed.

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8

Question

Let $G$ be a Group and suppose $a, b \in G$ are elements of $G$ such that $a b$ has finite order. Prove that:

| a b | = | b a |

Proof

Since $a b$ is finite order let $| a b | = n$ . Then $(a b)^{n} = 1$ . Now notice:

\begin{aligned} (a b)^{n} & = 1 \\ a^{n} b^{n} & = 1 \\ b^{n} & = a^{- n} \\ 1 & = b^{- n} a^{- n} \\ 1 & = (b a)^{- n} \\ (b a)^{n} & = 1 \end{aligned}

Thus $| b a | = n = | a b |$ .
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9 (Hard)

Question

Recall that 2-cycles in $S_{n}$ (see Symmetric Group & Permutations) are also called transpositions. We say a permutation $σ \in S_{n}$ is even if it can be written as a product of an even number of transpositions; otherwise we say it is odd. For example, in $S_{4}$ the 4-cycle $σ = (1 2 3 4)$ is odd because it can be written as a product of three transpositions, namely $σ = (1 4) (1 3) (1 2)$ .
For each permutation $σ \in S_{n}$ , define the sign of $σ$ as follows:

sgn (σ) = {\begin{cases} 1, & σ is even \\ - 1, & σ is odd \end{cases}

(It is a known fact that $sgn$ is well defined)
a. Compute the sign of each element in $S_{3}$ .
b. Let $A_{n} \subseteq S_{n}$ denote the subset of all even permutations. Show that $A_{n}$ is a subgroup of $S_{n}$ . This subgroup is called the alternating group on $n$ elements.

Proof

$σ$	$sgn$
$()$	1
$(1 2)$	-1
$(1 3)$	-1
$(2 3)$	-1
$(1 2 3) = (2 3) (1 3)$	1
$(1 3 2) = (2 3) (1 2)$	1
b. Let $n \in N$ , and let $A_{n} \subseteq S_{n}$ only contain even permutations. To show $A_{n} \subseteq S_{n}$ we have already shown it being a subset, so we need to show closure.

For closure under our binary operation, let $σ_{1}, σ_{2} \in A_{n}$ . Then $σ_{1}, σ_{2} \in S_{n}$ are both even permutations. Notice then that since they both are even permutations, then they have $2 k_{1}, 2 k_{2}$ individual transpositions in their product expansions separately. Say:

σ_{1} = ψ_{1} ψ_{2} \dots ψ_{2 k_{1}}, σ_{2} = ψ_{2 k_{1} + 1} \dots ψ_{2 k_{1} + 2 k_{2}}

Where $k_{1}, k_{2} \in N$ and $ψ_{i}$ is a transposition for each $0 < i \leq 2 k_{1} + 2 k_{2}$ . Now then:

σ_{1} σ_{2} = ψ_{1} \dots ψ_{2 k_{1}} ψ_{2 k_{1} + 1} \dots ψ_{2 k_{1} + 2 k_{2}}

Which has a total of $2 (k_{1} + k_{2})$ transpositions, an even number! Thus then $σ_{1} σ_{2}$ is even, and thus $σ_{1} σ_{2} \in A_{n}$ .

For the inverse being closed, let $σ \in A_{n}$ again. Use our $σ_{1}$ definition to define $σ$ , where we use $k$ instead of $k_{1}$ . Notice that:

σ^{- 1} = {(\prod_{i = 1}^{2 k} ψ_{i})}^{- 1} = \prod_{i = 1}^{2 k} (ψ_{i})^{- 1}

Now, since $(ψ_{i})^{- 1}$ is still a transposition, then $σ^{- 1}$ has a product of an even amount of transpositions, and thus $σ^{- 1} \in A_{n}$ .

Therefore $A_{n}$ is a subgroup of $S_{n}$ for all $n$ .

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