# 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 $H$ is required to be the identity of $G$. One might require only that $H$ have an identity element, not that it is the same as the identity in $G$. Show that if $H$ has an identity at all, then it is the identity in $G$ , so this definition would be equivalent to the one given.

Proof

Let $e \in G$ be the identity element of the group $G$;
let $e_{H} \in H$ be an element that behaves like the identity for for $H$;

Take any $c \in H$ (since $H \subset G$, we also have $c \in G$).
We know that $e_{H}c = ce_{H} = c$ since $c \in H$,
and also that $ec = ce = c$, since $c \in G$.

Equating the two, $e_{H}c = ec$, and a right-multiplication by $c^{-1}$ gives $e_{H} = e$, 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 $H$, but works differently for some $c \not\in H$!
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 $c \in H$, we can just take $e_{H} \in H$. We get that $e_{H}e_{H} = e_{H}$. Now multiply by $e_{H}^{-1}$ to conclude $e_{H} = e$.
The argument, in plain words: a faux-identity cannot exist, because it would square to itself, and only the actual identity does that. — Z