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Section 13.2 Discovery guide

Subsection An example of Cauchy's Theorem

Warning 13.2.1.

Make sure you have already read about words in the Pre-read section.

Recall that

\begin{equation*} \grporder{A_4} = {4! \over 2} = 12 \text{.} \end{equation*}

We saw in Discovery 11.7 that even though \(6\) divides evenly into \(12\text{,}\) \(A_4\) has no subgroup of order \(6\text{.}\)

However, Cauchy's Theorem says that \(A_4\) should have an element of order \(3\text{,}\) since \(3\) is a prime divisor of \(12\text{.}\) And this element of order \(3\) will generate a cyclic subgroup of order \(3\text{,}\) so it will turn out that \(A_4\) does have a subgroup of order \(3\text{.}\)

Let's go looking for this element of order \(3\text{.}\) In what follows, it won't actually be important that we are working with \(G = A_4\text{,}\) all that will matter is that we are working with a group \(G\) with \(\grporder{G} = 12\text{.}\)

Discovery 13.1.

We are looking for \(x\) in \(G\) so that \(x^3 = e\text{.}\) The left-hand side of this equality is a word with three letters:

\begin{equation*} x x x \text{.} \end{equation*}

Let's step back and consider all words of length three, where letters represent elements of \(G\text{:}\)

\begin{equation*} x_1 x_2 x_3 \text{.} \end{equation*}
(a)

How many such word of length three are possible?

Remember:

  • we can have repeat letters;

  • here we are only considering words, not the results of computing out those words using the group operation; and

  • we are assuming \(\grporder{G} = 12\text{.}\)

(b)

Of those words of length three, how many compute to \(e\text{?}\)

Hint.

Consider the equality

\begin{equation*} x_1 x_2 x_3 = e \text{.} \end{equation*}

Can you use group operations to rearrange it algebraically to say something about one “letter” in terms of the other two? How does this restrict the number of possibilities for that word on the left?

Discovery 13.2.

Use group algebra to demonstrate that if the word

\begin{equation*} x_1 x_2 x_3 \end{equation*}

computes to \(e\text{,}\) then so do the “cycled” words

\begin{equation*} x_2 x_3 x_1, \qquad x_3 x_1 x_2 \text{.} \end{equation*}

Careful: \(G\) might not be Abelian, so you can't just rearrange the order of multiplication; you will need to rearrange the equality

\begin{equation*} x_1 x_2 x_3 = e \text{.} \end{equation*}

Discovery 13.2 says that we can partition the collection of all three-letter words that compute to \(e\) into cells consisting of cycles:

\begin{gather} \eqclass{x_1 x_2 x_3} = \{ x_1 x_2 x_3, x_2 x_3 x_1, x_3 x_1 x_2 \} \text{.}\tag{✶} \end{gather}

From the way we've expressed this class, it appears that every class has three different words in it. (Again, not considering what group element these words compute to, because right now we are assuming they all compute to \(e\text{.}\)) But remember that a word could contain repeat letters. Indeed, remember that we are actually looking for a word \(x x x\) that computes to \(e\text{.}\)

Discovery 13.3.

(a)

Give an example of a three-letter word that computes to \(e\) and whose class (as in (✶)) actually only contains three repeats of the same word.

Note: Make sure you can justify the condition that the word must compute to \(e\text{,}\) or else you will likely just be assuming exactly that which we would ultimately like to prove in this subsection.

(b)

Is it possible that a class as in (✶) contains exactly two different words, with one repeat? Justify your answer.

Discovery 13.4.

In Task 13.1.b you counted all possible three-letter words that compute to \(e\text{;}\) write \(N\) to represent this number. Using the partition into classes of cycled words, we can count this total another way. Let \(n_1\) represent the number of classes that contain the same word repeated, let \(n_2\) represent the number of classes that contain exactly two different words and one repeat, and let \(n_3\) represent the number of classes that contain three different words. Then we have

\begin{gather} N = 1 \cdot n_1 + 2 \cdot n_2 + 3 \cdot n_3 \text{.}\tag{✶} \end{gather}

(Though you should be able to use Task 13.3.b to simplify the formula on the right.)

Task 13.3.a says that \(n_1 \ge 1\text{.}\) Substitute the actual number you computed for \(N\) in Task 13.1.b into (✶), and use this expression to come to the conclusion that \(n_1 = 1\) is not possible.

Hint.

Take \(n_1 = 1\) in (✶), and show that this leads to an impossible situation relative to your value for \(N\text{.}\) (Don't forget to use Task 13.3.b as well!)

Discovery 13.4 demonstrates that there must be more than one class that actually consists of just one repeated word. The repeated word in such classes must be of the form

\begin{equation*} x x x \text{,} \end{equation*}

which computes to \(x^3\text{.}\) Since there is more than one such word, there must exist an \(x \neq e\) with \(x^3 = e\text{,}\) as desired.

This completes the verification of Cauchy's Theorem in the special case of \(\grporder{G} = 12\) and \(p = 3\text{,}\) but the proof in the general case of \(\grporder{G} = n\) and prime \(p\) dividing \(n\) works exactly the same, only with cycles of words of length \(p\) that compute to \(e\text{,}\) instead of length \(3\text{.}\)

Warning 13.2.2.

In all of this, we were assuming that \(x^p = e\) implies that \(\grporder{x} = p\text{,}\) which is only valid because we were working with prime \(p\text{.}\) If \(p\) were not prime we could not make this leap of logic — see Warning 11.0.4.

Subsection Applying Cauchy's Theorem

Cauchy's Theorem helps us explore the structure of groups by guaranteeing that subgroups of certain sizes exist.

Discovery 13.5. Groups of order \(6\).

In this activity we will try to answer the following question: What possible forms can a group of order \(6\) take?

(a)

List the prime divisors of \(6\text{.}\) (Note: \(1\) is not prime.)

(b)

Cauchy says that \(G\) must have at least one element of each of the prime divisors of \(6\text{.}\) Write \(r\) for the one of larger order and \(s\) for the one of smaller order. Write out the elements of the two cosets \(H\) and \(H s\) for \(H = \gen{r}\text{.}\) This is now all the elements of \(G\text{,}\) partitioned into cosets. What group do these elements remind you of?

(c)

Let's see if the algebra of \(G\) has to work out as it does in that group you answered for the final question of the previous task. Two of the three algebraic conditions of that group are already met because they are the same as our assumptions about the orders of \(r\) and \(s\text{.}\) Group \(G\) has six elements, so there are six possibilities for what the result of computing \(s r\) could be. Use algebra and our assumptions about the orders of \(r\) and \(s\) to demonstrate that four of these six possibilities are actually impossible.

(d)

For the other two possibilities for what \(s r\) could be, one is precisely the third condition that would make the algebra of \(G\) exactly the same as the group you answered at the end of Task b (and so \(G\) is isomorphic to that group in this case).

What about the other possibility? Use this condition to prove that in this case

\begin{equation*} G \isoto \gen{r} \times \gen{s} \text{.} \end{equation*}

To what familiar group of order \(6\) does that make \(G\) isomorphic?

Discovery 13.6.

In each of the following cases, determine a complete list of familiar example groups so that in every case \(G\) must be isomorphic to one of the groups in your list.

(a)

\(\grporder{G} = 1\text{.}\)

(b)

\(\grporder{G} = 2\text{.}\)

(c)

\(\grporder{G} = 3\text{.}\)

(d)

\(\grporder{G} = 4\text{.}\)

(e)

\(\grporder{G} = 5\text{.}\)

(f)

\(\grporder{G} = 7\text{.}\)

Discovery 13.7.

Already the analysis for \(\grporder{G} = 8\) gets fairly involved — you can read about it in the textbook. For now, make a list of four example groups of order \(8\text{,}\) where no two groups in your list are isomorphic.

Hint.

Likely three of your four examples will be Abelian. Don't forget that we have a way to build a “larger” group out of smaller ones — see Section 10.1.