** Using WeBWorK **

## DBsubject(Abstract algebra)
## DBchapter(Groups)
## DBsection(Subgroups)
## Institution(University of Lethbridge)
## Author(Joy Morris)
## Level(4)
## MO(1)
## KEYWORDS('subgroup proof')




DOCUMENT();       

loadMacros("PGessaymacros.pl","PGML.pl");

TEXT(beginproblem());
$showPartialCorrectAnswers = 0;


$choiceArray = ( 0,0,0,0,0);

$rand = random (0,4,1);

$choiceArray[$rand] = 1;

$step1[0] = String("Define the set. `H = \{ a^i | 1 \le i \le 5\}`.");

$step1[1] = "Define the set. `H = \{ a^i | 0 \le i \le 5\}`.";

$step2[0] = "Show `e \in H`. Notice that since `|a|=5` we have `a^5=e`. Therefore `e=a^5` and since `1 \le 5 \le 5` we see that `e \in H` by our definition of `H`.";

$step2[1] = "Show `e \in H`. We see that `e=a \in H` so `e \in H` by our definition of `H`.";

$step3[0] = "Let `x,y` be arbitrary elements of `H`. By definition of `H`, this means that there exist `i,j` with `1 \le i,j \le 5` such that `x=a^i` and `y=a^j`.";

$step3[1] = "Let `x,y` be arbitrary elements of `H`. By definition of `H`, this means that there exist `i,j` with `1 \le i,j \le 5` and `i \neq j` such that `x=a^i` and `y=a^j`.";

$step4[0] = "Deduce that `x^{-1} \in H`. If `i=5` so that `x=a^5=e` then `x^{-1}=e=a^5 \in H` by definition of identity and inverses, since `e \cdot e = e`. Otherwise, `1 \le i \le 4` and we have `x^{-1}=(a^i)^{-1}=a^{-i}`, and since `1 \le i \le 4` we have `-4 \le -i\le -1`. Since `a^5=e`, we have `x^{-1}=a^{-i}=e\cdot a^{-i}=a^{5-i}`, and `1 \le 5-i \le 4`, so again `x^{-1} \in H`.";

$step4[1] = "Deduce that `x^{-1} \in H`. If `i=5` so that `x=a^5=e` then `x^{-1}=e=a^5 \in H` by definition of identity and inverses, since `e \cdot e = e`. Otherwise, `1 \le i \le 4` and we have `x^{-1}=(a^{-i})^{-1}=a^{i}`, and since `1 \le i \le 4` we see that `x^{-1} \in H`.";

$step5[0] = "Deduce that `xy \in H`. We have `xy=a^ia^j=a^{i+j}`. If `1 \le i+j \le 5` then it is immediately clear that `xy \in H`. If `i +j \ge 6` then since `1 \le i,j \le 5` we have `6 \le i+j \le 10`. So `a^{i+j}=a^{5+k}=a^5a^k=a^k` for some `1 \le k \le 5`. Therefore, again we see that `x=a^k \in H`.";

$step5[1] = "Deduce that `xy \in H`. We have `xy=a^ia^j=a^{i+j}`. If `1 \le i+j \le 4` then it is immediately clear that `xy \in H`. If `i +j \ge 5` then since `1 \le i,j \le 4` we have `5 \le i+j \le 8`. So `a^{i+j}=a^{4+k}=a^4a^k=a^k` for some `1 \le k \le 5`. Therefore, again we see that `x=a^k \in H`.";
                       
BEGIN_PGML

Consider the following subgroup proof. Read it carefully. In the box below, write about any errors you find in this proof and explain how to fix them. If you need to use mathematical notation, just do your best to type in what you mean.

Let [` G `] be a group, and let [` a\in G `] with [`|a|=5`]. We will prove that [`H=\{ a,a^2,a^3,a^4,a^5=e\}`] is a subgroup of [`G`].

[$step1[$choiceArray[0]]]

[$step2[$choiceArray[1]]]

[$step3[$choiceArray[2]]]

[$step4[$choiceArray[3]]]

[$step5[$choiceArray[4]]]

By the five-step subgroup test, we have proven that [`H \le G`].

[@ essay_box(5,50) @]*
END_PGML

ANS(essay_cmp());

ENDDOCUMENT();