Math 441 - Abstract Algebra - Schedule

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In class, someone asked if it was enough to just show that something worked as an inverse on one side. In linear algebra, you showed this was true for matrices, namely that if you knew $BA=I$, then you also know that $AB=I$. This fact is also true in groups.

Problem.One Sided Inverses Are Sufficient

Let $G$ be a group and let $x,y\in G$. Prove that if $yx=e$, then $y$ is the inverse of $x$. In other words, if you can show that $y$ is a one-sided inverse of $x$, then it must be a two-sided inverse.

We need to review the definition of the order of a group.

Definition ($|a|$ and $|G|$ - Order For Elements and Groups)

Let $G$ be a group with identity $e$, and let $g\in G$.

The $\textdef{order}$ of $G$, denoted $|G|$, is the cardinality of $G$.
The $\textdef{order}$ of $g$, denoted $|g|$, is the smallest positive integer $n$ such that $g^n = e$, if such an $n$ exists. If no such $n$ exists, we say $g$ has infinite order.

If you feel like you need extra practice with this definition, then please complete the following exercise.

Exercise.Practice with Order

In each part below, you are given an element $a$ of a group $G$. Find the order of that element.

Let $a= R_{270}$ which is an element of $D_{8}$, the automorphisms of a square.
Let $a= R_{30}$ which is a 30 degree rotation and an element of $D_{48}$, the automorphisms of a regular 24-gon.
Let $a=\left<\begin{bmatrix}2&1\\1&0\end{bmatrix}\right>$ in the general linear group $GL(2,\mathbb{Z}_3)$. [Hint: Just as all the previous problems, start computing powers of this matrix until you obtain the identity.]
Let $a=(1,2,3,4,5)$ which is an element of the set of all permutations of $X=\{1,2,3,4,5\}$.
Let $a=3$ as an element of $U(7)$.
Let $a=2$ as an element of $U(17)$.
Let $a=3$ as an element of $U(17)$.

Click to see a solution.

In each instance, we just have to compute $a^k$ for $k=1,2,3,\ldots$ until $a^k=e$. The smallest $k$ for which this occurs is the order of $a$. You should have obtained the answers 4, 12, 8, 5, 6, 8, 16. Make sure you can explain why for each.

Exercise.

Suppose we know that $a^6=e$.

Explain why this is not enough information to state the order of $a$. (Look at the definition. What are we missing?)
In addition to knowing that $a^6=e$, someone else notices that $a^4=e$. Prove that the order of $a$ cannot be 4. In particular show that $a^2=e$, so the order of $a$ is either $2$ or $1$.

Click to see a solution.

The order of an element is the SMALLEST positive integers $n$ such that $a^n=e$. If all we know is that $a^6=e$, then the order might be 6, or some number less.
If we know that both $a^6=e$ and $a^4=e$, then since $6=4+2$ (the division algorithm), we know that $e=a^6=a^{4+2}=a^4a^2=ea^2=a^2$. This shows that $e=a^2$, which means the order now can at most be 2.

Problem (The last half of what Kinsey shared in class - THIS IS A KEY PROBLEM)

Suppose that $a$ is an element of order $n$. Prove that $a^i=a^j$ if and only if $i-j$ is a multiple of $n$.

Problems from the past

We still have the following problems left that we need to present in class.

Monday's, The Center Of A Dihedral Group
All four problems from Wed. I'll be reassigning presenters for these, based on your preparation logs. Please report your progress on all 7 problems for today.

Please rework on these problems.

Click to view these problems, as well as some hints.

Problem 61 (The Center Of A Dihedral Group)

Let $G=D_4$, the automorphism group of the square. Recall that $Z(G)$ is the center of the group, or the set of elements that commute with every element of the group.

What is $\langle R_{90} \rangle$? What is $\langle R_{180} \rangle$? What is $\langle R_{270} \rangle$? What is $\langle H \rangle$?
Does $R_{90}\in Z(G)$? Explain. (Does $R_{90}$ commute with every element in $G$? In particular, does $R_{90}H=HR_{90}$?)
Compute the center $Z(G)$ and show that it consists of more than just $R_0$. Make sure you can explain why each element is either in $Z(G)$, or not in $Z(G)$.

Hint: There are several ways to proceed. One common approach is to just, by brute force, compute the product of every two elements. You should find there are two elements in the center.

Here's a faster way. If you rotate clockwise $x$ degrees and then flip, that's the same as first flipping and then rotating $x$ degrees counterclockwise. The only way you could end up with a rotation and a flip commuting is if a counter clockwise rotation of $x$ degrees and a clockwise rotation of $x$ degrees resulted in the same action. You can also read the solution in chapter 3 of your text on page 63.

Problem 53 (Finite Subgroup Test)

Let $G$ be a group. Suppose that $H$ is a nonempty finite subset of $G$ and that $H$ is closed under the operation of $G$ (so if $a,b\in H$, then we must have $ab\in H$). Prove that $H$ is a subgroup of $G$.

Hint. If you use the subgroup test (show that $H$ is a nonempty subset of $G$ that is closed under the operation and taking inverses), then we get to assume it's a nonempty subset of $G$ that is closed under the operation. All you have to do is explain why it's closed under taking inverses. The work you did in the problem The Inverse In A Finite Group Is A Power Of The Element should help you quite a bit. However, you can't use this theorem directly because you do not know that $G$ is a finite group. You'll want to reuse your work from that problem, not refer to the problem.

Problem 62 ($\langle a^k\rangle = \langle a^{\gcd(k,|a|)}\rangle$)

Let $a$ be an element of order $n$ and let $k\in\mathbb{N}$. Prove that $\langle a^k\rangle = \langle a^{\gcd(k,n)}\rangle$.

You have to show that each set is a subset of the other. If you are completely stuck, then work with an example. Try letting $G=H_12$, the set of simple shift permutations on 12 letters. Pick an element, like $\phi_2$ in this set, where the order of $\phi_2$ is clearly $n=6$. Let $k=4$, so that $d=\gcd(4,6)=2$. Can you now prove that $\left<\phi_2^k\right>\subseteq \left<\phi_2^{\gcd(k,n)}\right>$ and that $\left<\phi_2^{\gcd(k,n)}\right>\subseteq \left<\phi_2^k\right>$. You can list out the elements of both sets to see that the fact is true, however don't let your proof be, "I just wrote down the elements, so clearly it's true" because that won't generalize. Instead, you've got to try to prove the result without referring to your list of elements. What connection is there between $k=4$ and $d=2$ that will get you each set inclusion? You'll need the fact that $d$ is a divisor of $k$ for one proof, and you'll need the fact that since $d=\gcd(k,n)$ we know that $d=sk+tn$ for some $s,t\in \mathbb{Z}$ for the other proof. You'll also need to at some point realize that $\phi_2^n=e$, as $n$ is the order of $\phi_2$.

You can read the general proof in your text in chapter 4.

Problem 63 ($|a^k| = |a|/\gcd(k,|a|)$)

Let $a$ be an element of order $n$.

If $d$ is a divisor of $n$, then prove that $|a^d|=n/d$.
For any $k\in \mathbb{N}$, prove that $|a^k| = n/\gcd(k,n)$.

You'll need to use the definition of order, and two previous problems to make short work of this problem. In particular, if you pay attention to the fact that the order of an element always matches the order of the subgroup generated by that element, then you can make sense of $|a^k|$ in terms of a greatest common divisor.

What is $(a^d)^{n/d}$? If $j<n/d$, why is $(a^d)^{j}\neq e$?

From the previous problem, we know that $\left<a^k\right>=\left<a^{\gcd(k,n)}\right>$. Why do we know that $|a^k| = |a^{\gcd(k,n)}|$? Why does part 1 then finish part 2 of this problem?

Problem 64 ($\langle a^i \rangle = \langle a^j \rangle$ iff $\gcd(i,n)=\gcd(j,n)$)

Let $G$ be a group. Suppose that $a\in G$ has order $n>1$. Prove the following two facts:

If $G$ is a cyclic group, then the order of $a$ divides the order of the group.
We have $\langle a^i \rangle = \langle a^j \rangle$ if and only if $\gcd(i,n)=\gcd(j,n)$.

Click for a hint.

If $G$ is cyclic, then pick a generator, say $b$. This means $G=\langle b\rangle$. Why does this mean $a=b^k$ for some $k$? Use this to show that $b$ has finite order, say $m$. Then use problem 63 with $b^k$, where $|b|=m$, rather than $a^k$ and $|a|=n$.
Problem 63 will get you one direction, and problems 62 will get you the other.

The second fact proves the following three corollaries by letting $i=1$.

We have $\langle a \rangle = \langle a^j \rangle$ if and only if $\gcd(n,j)=1$.
A simple shift permutation $\phi_j$ on $n$ letters generates all simple shift permutations on $n$ letters if and only if $\gcd(n,j)=1$.
An integer $j\in \mathbb{Z}_n$ is a generator of $\mathbb{Z}_n$ if and only if $\gcd(n,j)=1$.

What does this have to do with the problem When Does An Integer Have A Modular Multiplicative Inverse? You should see a connection between this problem and elements of $U(n)$.

If the group is cyclic, then what does the previous problem say about the order of $a^k$ for any $k$.

For the iff proof, remember to prove both directions. The fact that $\left<a^k\right>=\left<a^{\gcd(k,n)}\right>$ should help you quickly show that two sets are equal. The fact that $|a^k|=n/\gcd(k,n)$ should help you prove two numbers are the same.

For more problems, see AllProblems

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