Math 360, Fall 2013, Assignment 7

Mathematicians are like Frenchmen: whatever you say to them they translate into their own language, and forthwith it is something entirely different.

- Goethe

Carefully define the following terms, then give one example and one non-example of each:

1. Left congruence (i.e. the relation $\sim_L$ discussed on page 97 of the text).
2. Right congruence.
3. Left coset.
4. Right coset.
5. Index of a subgroup.

Carefully state the following theorems (you need not prove them):

1. Lagrange's Theorem.
2. Theorem concerning the indices of nested subgroups (Theorem 10.14 in the text).

Solve the following problems:

1. Section 10, problems 1, 7, 13, 15, 22, and 24.

--------------------End of assignment--------------------

Solutions:

Definitions

1. Left Congruence ($\sim_{L,H}$)

   Fix any group G, with a subgroup H. Define the relation, denoted $\sim_{L,H}$:
   $a\sim_{L,H}b \Leftrightarrow b^-1a\in H$
   We call this relation left congruence modulo H. Note that this is always an equivalence relation.(Proof)
   Example: Let $G=S_3, H=A_3$. $(1,2,3) \sim_{L,H} (23)(21)$, since $(21)(23)(123)=(132)\in H$
   Non-example: Let G and H be defined as above. $(1)(2)(3)\ \neg \sim_{L,H} (13)$, since $(13)(1)(2)(3)=(13) \notin H$.

2. Right Congruence

   Definition:

   Let \(G\) be a group and \(H\) be a subgroup of \(G\). Then for any elements \(a,b\in G\), \(a\) is right congruent modulo \(H\) if and only if:$$ ab^{-1}\in H $$

   Right congruence modulo \(H\) is an equivalence relation.

   Example:

   Let \(G = \mathbb{Z}_6\) and \(H=\langle 2\rangle\). Then \(2\) is right congruent to \(4\) modulo \(\langle 2\rangle\) because \(2 +_6 (-4) = 4\in \langle 2\rangle\).

   Non-Example:

   Taking the same structures, \(2\) is not right congruent to \(3\), because \(2+_6(-3) = 5 \not\in\langle2\rangle\).

3. Left Coset

   Definition:

   Take a group \(G\), a subgroup of \(G\) called \(H\), and an element \(a\in G\). The left coset of \(H\) containing \(a\) is the set:$$ aH = \{ah|h\in H\} $$

   In other words, its the subset of \(G\) made up of \(a\) multiplied (operated, really) by all the elements of \(H\).

   Example:

   Let \(G = \mathbb{Z}_6\) and \(H = \langle 2 \rangle\). The left coset of \(H\) containing \(3\) is then:$$\begin{eqnarray*} 3+0 &=& 3\\ 3+2 &=& 5\\ 3+4 &=& 1\\ 3H &=& \{1,3,5\} \end{eqnarray*} $$

   Non-Example:

   A decent non-example is hard - it relies on the distinction between a left and a right coset, and left and right cosets are only distinct in non-abelian groups.

4. Right Coset

   Definition:

   Take a group \(G\), a subgroup of \(G\) called \(H\), and an element \(a\in G\). The right coset of \(H\) containing \(a\) is the set:$$ Ha = \{ha|h\in H\} $$

   In other words, its the subset of \(G\) made up of \(a\) multiplied (operated, really) by all the elements of \(H\).

   Example:

   Let \(G = \mathbb{Z}_6\) and \(H = \langle 2 \rangle\). The left coset of \(H\) containing \(3\) is then:$$\begin{eqnarray*} 0+3 &=& 3\\ 2+3 &=& 5\\ 4+3 &=& 1\\ H3 &=& \{1,3,5\} \end{eqnarray*} $$

   Non-Example:

   Good non-examples are difficult, as before.

5. Index of a Subgroup

   Definition:

   Given a group \(G\) and a subgroup \(H\), the index of \(H\) is the number of left cosets of \(H\).

   Example:

   Let \(G=\mathbb{Z}_6\) and \(H=\langle 3\rangle\). The index of \(H\) is 3.

   Non-Example:

Theorems

1. Lagrange's Theorem

   If \(H\) is a subgroup of \(G\), then \(|H|\) divides \(|G|\).

2. Theorem Concerning The Indices of Nested Subgroups

   Let \(G\) be a group, \(H\) a subgroup of \(G\), and \(K\) are subgroup of \(H\) (\(K\leq H \leq G\)). If \((H:K)\) (the index of \(K\) in \(H\) and \((G:H)\) are both finite, then \((G:K)\) is finite, and \((G:K)) = (H:K)*(G:H)\).

Book Solutions

1. 10.1

   The cosets of \(4\mathbb{Z}\) are: \([0],[1],[2],[3]\) - equivalence classes modulo \(4\).

2. 10.7

   The cosets are not the same (the group is not commutative). The actual cosets are:$$ \begin{eqnarray*} \rho_0 &=& \{\rho_0,\mu_2\}\\ \rho_1 &=& \{\rho_1, \delta_1\}\\ \rho_2 &=& \{\rho_2, \mu_1\}\\ \rho_3 &=& \{\rho_3, \delta_2\}\\ \mu_1 &=& \{\mu_1, \rho_2\}\\ \mu_2 &=& \{\mu_2, \rho_0\}\\ \delta_1 &=& \{\delta_1,\rho_1\}\\ \delta_2 &=& \{\delta_2,\rho_3 \} \end{eqnarray*} $$