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
4. Right Coset
5. Index of a Subgroup