This is Revision (30 July 2016) of the Smooth Infinitesimal Analysis FAQ.

# The first year of QE’D

In Spring 2015, Totally QE’D read **Algebra, Second Edition** by Michael Artin, following the Math E222 Abstract Algebra Harvard Extension Course. We’ve been asked for pdf files with solutions to the problem sets. They will appear here as they’re completed.

#### Group Theory

Problem Set 7: First Isomorphism Theorem

Problem Set 6: Chinese Remainder Theorem, Casting out Nines

Problem Set 5: Fibers, Conjugacy, Normal Subgroups

Problem Set 4: Klein 4-group, Inner Automorphisms

Problem Set 3: Subgroups of the Integers, Cyclic Groups, Group Homomorphisms

Problem Set 2: Lagrange’s Theorem, Laws of Composition

Problem Set 1: Matrices

#### Modules and Vector Spaces

Problem Set 11 coming soon

Problem Set 10: Linear Groups, Change of Basis as Conjugation

Problem Set 9: Linear Independence in Function Spaces, Subspace, Dimension

Problem Set 8: Field Homomorphisms, Wilson’s Theorem

# Problem 2.2.15

We had problems explaining to you why this problem makes sense, so here’s a write-up. In the definition of subgroup, the identity element in is required to be the identity of . One might require only that have an identity element, not that it is the same as the identity in . Show that if has an identity at all, then it is the identity in , so this definition would be equivalent to the one given.

** Proof **

Let be the identity element of the group ;

let be an element that behaves like the identity for for ;

Take any (since , we also have ).

We know that since ,

and also that , since .

Equating the two, , and a right-multiplication by gives , **Qed**.

The argument could also be: since the identity element is unique in any group, and since a subgroup consists only of the elements of the “parent” group, the identity elements of the two groups must coincide.

— M

One has to be careful with that second argument! Sure, there is only one identity element in every group, but that doesn’t mean (in itself) that there’s can’t be a faux-identity, which works like the identity for elements of a subgroup , but works differently for some !

In fact, faux-identities can’t exist, and the argument above establishes that.

If we read the proof carefully, we’ll realise that it can be simplified even further! Instead of considering an arbitrary , we can just take . We get that . Now multiply by to conclude .

The argument, in plain words: a faux-identity cannot exist, because it would square to itself, and only the actual identity does that.— Z

# Hello world!

Totally QE’D is an undergraduate Mathematics reading group at the University of Stirling. In 2015, we’ll be reading **Algebra, Second Edition** by Michael Artin,, following along with the Harvard Math E222 Abstract Algebra course.

Your hosts/instructors are: Benjamin Merlin Bumpus, Zoltan Attila Parades and Pernille Rytterhuus.