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Because of how we defined groups, anytime we span a collection of permutations of a set $X$ we should get a group. The next problem has you show this. Each question can be answered by referring to things we have already shown. Feel free to look at the pages September 2017 and October 2017 for a list of all the problems we have solved each day. If you have not yet printed the work we are doing, I would suggest that you print a copy for reference as you work on new problems.

Problem 45 (Spans Of Permutations Are Subgroups)

Let $X=\{1,2,3,\ldots,n\}$. Let $S_n$ be the set of all permutations of $X$ (so we know there are $n!$ such permutations).

  1. Show that $S_n$ is a group under function composition $\circ$.
  2. If $H$ is a subset of $S_n$ that is closed (under composition combinations of permutations), prove that $H$ is a group under function composition.
  3. Let $S$ be a nonempty subset of $S_n$. Show that $\text{span}(S)$ is a group under function composition.


In the previous problem, we saw if $S\subseteq S_n$ then $\text{span}(S)$ is a not only just a subset of the group $S_n$, but is itself a group. The operation we use in the smaller group $(\text{span}(S),\circ)$ comes from the operation we used in the larger group $(S_n,\circ)$. Because of this close connection, we call $\text{span}(S)$ a subgroup of $S_n$ and write $\text{span}(S)\leq S_n$. Let's now make a formal definition.

Definition (Subgroup)

Let $(G,\cdot)$ be a group, and let $H$ be a nonempty subset of $G$. Then $H$ is called a $\textdef{subgroup}$ of $G$ if the following hold:

  1. $\textbf{[Closure]}$ for all $h,k\in H$ we have $h\cdot k\in H$, and
  2. $(H,\cdot)$ is a group.

When $H$ is a subgroup of $G$, we write $H\le G$. Any subgroup of $G$ that is not equal to $G$ itself we call a $\textdef{proper subgroup}$. The subset of $G$ consisting of just the identity we call the $\textdef{trivial subgroup}$.


In the proof of problem Characterizing Closed Sets Of Permutations, we did not need to use the first and fourth properties listed to show that the set of permutations $H$ was closed. The next problem replicates this, but now in terms of subgroups instead of in terms of closed sets of permutations.

Problem 46 (The Subgroup Test - Subgroups Are Subsets That Are Closed Under Products And Taking Inverses)

Suppose that $G$ is a
group
Definition (Group)

Let $G$ be a nonempty set, and let $*$ be a binary operation on $G$, which means for every $x,y\in G$ we have $x*y\in G$ $\textbf{[Closure]}$. The structure $\mathbb{G} = (G,*)$ is called a $\textdef{group}$ if the following hold.

  1. $\textbf{[Associativity]}$ For all $x,y,z\in G$ we have $(x* y)* z = x* (y* z)$.
  2. $\textbf{[Identity]}$ There is an $e\in G$ such that for all $x\in G$ we have $x * e = e* x = x$.
  3. $\textbf{[Inverses]}$ For all $x\in G$ there is a $y\in G$ such that $x* y = y* x = e$.

We usually simply write $G$ when referring to the entire structure $\mathbb{G}=(G,*)$. The element $e$ from the second point is called the $\textdef{identity}$. The element $y$ from the third point is called the $\textdef{inverse}$ of $x$ and is usually denoted $x^{-1}$. One often simply writes $xy$ in place of $x*y$, and for every positive integer $n$, we'll write $x^n$ as shorthand for $x* x* \cdots * x$ ($n$ times).

and $H$ is a nonempty subset of $G$ that satisfies the following properites:
  1. If $a,b\in H$, then $ab\in H$. (We say $H$ is closed under the group operation.)
  2. If $a\in H$, then $a^{-1}\in H$. (We say $H$ is closed under taking inverses.)
Prove that $H$ is a
subgroup
Definition (Subgroup)

Let $(G,\cdot)$ be a group, and let $H$ be a nonempty subset of $G$. Then $H$ is called a $\textdef{subgroup}$ of $G$ if the following hold:

  1. $\textbf{[Closure]}$ for all $h,k\in H$ we have $h\cdot k\in H$, and
  2. $(H,\cdot)$ is a group.

When $H$ is a subgroup of $G$, we write $H\le G$. Any subgroup of $G$ that is not equal to $G$ itself we call a $\textdef{proper subgroup}$. The subset of $G$ consisting of just the identity we call the $\textdef{trivial subgroup}$.

of $G$.


Problem 50 (The Inverse In A Finite Group Is A Power Of The Element)

Let $G$ be finite group with $a\in G$. Prove that there exists a positive integer $k$ such that $a^k=a^{-1}$.

Hint: Consider the sequence $a, a^2, a^3, a^4, \ldots$. Why must you eventually have a repeated element? So if $a^i=a^j$ for some $j>i$, how can you use this to find $a^{-1}$?

Problem 51 (Inverses In Groups)

Suppose that $G$ is a group with $a,b\in G$.

  1. Prove that the inverse of $a^{-1}$ is $a$.
  2. Prove that the inverse of $ab$ is $b^{-1}a^{-1}$.
  3. If $a_1,a_2,a_3,\ldots, a_n\in G$, state the inverse of $a_1a_2a_3\cdots a_n$. Use induction to prove your claim.
If you see yourself repeating an induction proof similar to what we've been doing, then you're on the right track.

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 54 (The Intersection Of Two Subgroups)

Suppose that $G$ is a group and that $H$ and $K$ are subgroups of $G$. Prove that $H\cap K$ is a subgroup of $G$.


Problem 55 (The Union Of Two Subgroups)

Suppose that $G$ is a group and that $H$ and $K$ are subgroups of $G$. Is $H\cup K$ a subgroup of $G$? Either prove that is, or find a counterexample.


Definition (Abelian Group)

Let $G$ be a group. If $ab=ba$ for every $a,b\in G$ (so the group operation is commutative), then we say that $G$ is Abelian.

Definition ($Z(G)$ - Center Of A Group)

Let $G$ be a group. The center of the group, written $Z(G)$, is the set of elements $x\in G$ that commute with every element of $G$, which we can write symbolically as $$Z(G)=\{x\in G\mid gx=xg \text{ for all } g\in G\}.$$

The $Z$ comes from the german word "Zentrum" (see Wikipedia).

Problem 56 (The Center Of Group Is A Subgroup)

Prove that the center $Z(G)$ of a group $G$ is a subgroup of $G$. If $G$ is Abelian, then what is $Z(G)$?


Problem 57 (Powers Of Products In An Abelian Group)

Suppose $G$ is an Abelian group. Prove that if $a,b\in G$, then $(ab)^2=a^2b^2$. Then use induction to prove that if $a,b\in G$, then $(ab)^n=a^nb^n$ for each $n\in \mathbb{N}$.



For more problems, see AllProblems