Quotient Groups The elements of a group $G$ can be partitioned, using a subgroup $H$ of $G$, into cells called cosets. One of the nicest features of ESG will allow you to investigate these cosets, which sometimes form a new group called a factor or quotient group of $G$. Before the Lab You will prove the following theorem for homework: Theorem 6.1: Let $G$ be a group and $H$ a subgroup of $G$. Define $\sim $ by saying $a\sim g$ if and only if $ag^{-1}\in H$. Then $\sim $ is an equivalence relation. We would obtain a similar result if we defined $\sim $ by saying $a\sim g$ if and only if $g^{-1}a\in H$. In the the theorem, we have $a=hg$, for some $h\in H.$ In fact, as we let $h$ run through all the elements of $H$, we obtain all elements of $G$ that are related to $g$ under the equivalence relation $\sim $. The cell of the partition created by $\sim $ that contains $g$ is denoted by $Hg$ and is called the right coset of $H$ in $G$ that contains $g$. In the second case, we create $gH$, the left coset of $H$ in $G$ that contains $g$. Why do both $Hg$ and $gH$ contain $g$? For the two examples below, we'll consider the subgroups $H=\{1,r_1,r_2\}$ and $K=\{1,m_1\}$ of $S_3$. We'll compute the right and left cosets for each subgroup. Confirm the computations as you read through the example. Example 1:
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There are two distinct right cosets, $\{1,r_1,r_2\}$ and $\{m_1,m_2,m_3\}$, which form a partition of the elements of $G$.
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There are two distinct left cosets, $\{1,r_1,r_2\}$ and $\{m_1,m_2,m_3\}$. These agree with the right cosets: $Hg=gH$ for every $g\in G$. Example 2:
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There are three distinct right cosets, $\{1,m_1\}$, $\{r_1,m_2\}$, and $\{r_2,m_3\}.$ There are three distinct left cosets, $\{1,m_1\}$, $\{r_1,m_3\} $, and $\{r_2,m_2\}$. This time, however, not all of the left and right cosets agree. That is, there are elements $g\in S_3$ so that $Kg\neq gK.$ The distinction between $H$ and $K$ is crucial and at the heart of studying quotient groups. The collection of cosets forms a new group precisely when the left and right cosets agree. The problem, it turns out, is in trying to define a binary operation on the cosets. The coset operation is best defined through an example. We only have the operation defined on $G$ to use. Let's start by using the left cosets of $H$. Set $X=\{1,r_1,r_2\}$ and $Y=\{m_1,m_2,m_3\}$. Choose an element from $X $ and one from $Y,$ for example, $r_1\in X$ and $m_2\in Y$. Since $r_1m_2=m_1 $, and $m_1\in Y$, we define the coset operation $*$ by $X*Y=Y.$ No matter which representative of the cosets $X$ and $Y$ we choose, we will obtain the same answer in this case. So we can construct the Cayley table using the operation $*$:
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Check that this forms a group of order 2. Which coset serves as the identity element? If we try to do this with the left cosets of $K,$ however, we run into problems. Set $U=\{1,m_1\}$, $V=\{r_1,m_3\}$, $W=\{r_2,m_2\}.$ If we compute $V*W$ using the elements $m_3$ and $r_2$, we have $V*W=W.$ But if we do the calculation with $r_1$ and $m_2$, we obtain $V*W=U$. Obviously there is a problem here---the operation is not well-defined. In the lab, we will see how ESG demonstrates this with color. To summarize, we have the theorem: Theorem 6.2: Let $G$ be a group and let $H$ be a subgroup of $G$. The cosets of $H$ in $G$ form a group if and only if $Hg=gH$ for all $g\in G. $ When the condition of the theorem holds, we will say that the subgroup $H$ is a normal subgroup of $G$. The collection of cosets (right or left, of course) is then called the quotient or factor group, $G$ mod $H$, denoted by $G/H$. In our examples above, the subgroup $H$ is normal but $K$ is not. It is especially important for you to work out one example carefully on your own before you use the computer. Do all the computations by hand for question 1. 1. Consider the subgroups MATH and $H_3=\{1,r_2\}$ of $D_4.$ Find all the left cosets of the subgroups in $D_4$. Find all the right cosets of the subgroups in $D_4$. For which subgroups are the left and right cosets equal? For each subgroup identified in part (c), construct a group table for the quotient group $G/H_i.$ What familiar group has the same group table? Be sure to bring your subgroup lattices to the lab. In the Lab Check your answers to question 1 with ESG before continuing with the problems below. Choose option 3 (Subgroups and Cosets/Quotients) from the Group Properties Menu for $D_4$, and generate each subgroup $H_i$. Answer ``Y'' to the question, ``Would you like to see the left cosets of this subgroup?'' Look at the coloring of the Cayley table of $D_4,$ grouped by the left cosets of each subgroup. In some cases, you will be asked, ``Would you like to see the quotient table?'' Be sure you understand how the Cayley table is transformed when you answer ``Y.'' 2. In your own words, explain how you can determine from the table on the computer screen that the coset operation is well-defined or not well-defined. For problems 3-10, you may use the computer for your computations. Answer the following questions for each group $G$ and all nontrivial proper subgroups $H$ of $G$. Record the distinct left and right cosets of the subgroup $H$ in $G$. Is the subgroup $H$ normal in $G$? If so, have the computer construct a group table for the new quotient group $G/H$. What familiar group has the same group table? Your answer should be a known group from the ESG library. Be sure to record on your subgroup lattices which subgroups are normal. Note that there are no general formulas which help us figure out which group $G/H$ actually is. You have to rely on computing the order of $G/H$ and on knowing something about the various groups of that order. 3. $G=D_5$ 4. $G=D_6$ 5. $G=A_4$ 6. $G=Q_6$ 7. $G=D_7$ 8. $G=M$ 9. $G=D_8$ 10. $G=Q_8$ Further Work 11. Make at least two conjectures about the kinds of subgroups which always seem to be normal in a finite group $G$. (Hint: think about the special subgroups that we considered in an earlier lab or the index of the subgroup). 12. Make at least two conjectures about the factor groups $D_n/H$, where $H$ is either the commutator subgroup or center of $D_n$. 13. Prove Theorem 6.1: Let $G$ be a group and $H$ a subgroup of $G$. Define $\sim $ by $a\sim g$ if and only if $ag^{-1}\in H$. Then $\sim $ is an equivalence relation. This document created by Scientific WorkPlace 4.0.