Tag Archives: Group Theory

Assignment Result - Groups

December 24, 2021 Adrian

My second assignment on groups was marked and returned. As predicted, there's a lot to take away from this.

Basic Groups

At this level it seems it's not sufficient to use notation like $pH$ , $qH$ and $rH$ to refer to subgroups. In my head, I know what the binary operator of these subgroups is but for the benefit of the reader the (better) convention is to be explicit with the binary operator: $p+H$ , $q+H$ , and $r+H$ specifically.

Definition check: a cyclic group is just a group that contains a generator that generates the whole group. It doesn't have to "cycle" back to the beginning of the ordered set. eg, infinite sets can form a cyclic group under addition.

Mistakes in basic proofs. This one is a classic. With $p,q\in \mathbb{Q}$ , I had to prove $\phi(q)=\frac{3q}{2}$ was onto. My answer:

$\forall\: q\in\mathbb{Q},\: \exists\: p\in\mathbb{Q},\: \text{s.t.}\: \phi(p)=q,\: \therefore \phi$

Which I was all smug about because it looks pretty. But this is just the definition of the function being onto and I'd neglected to state the actual $p$ that always exists. Namely $p=\frac{2q}{3}$ .

Group Classification

Group decompositions can't be written as $\mathbb{Z}_{6} \times \mathbb{Z}_{45}$ . It needs to be written as $\mathbb{Z}_{3} \times \mathbb{Z}_{6} \times \mathbb{Z}_{15}$ . 3 and 15 are not coprime, so cannot be combined.

Finite Groups

As a part of a group presentation, there are kind of identities like $sr=r^{5}s$ . However, for the most part in your answer, especially when you're talking directly about elements being in your group, they must always be in the form $r^{i}s^{j}$ (with $r$ first).

In being asked to deduce an isomorphism, I heavy-handedly defined both groups (as they were small) and then also stated a function explicitly that would define the isomorphism. Apparently it's just enough to state that both groups were of order 2. Therefore they're isomorphic.

I need to become a lot more familiar with the concept of centres of a group. To show that a subgroup is the centre of its parent group, I needed to show that the subgroups elements were centres of the subgroup. I was was just trying to wing it, and although I essentially got the answer right, I showed I had a complete misunderstanding of centres.

The Last Group Theory Feedback

March 21, 2017 Adrian Leave a comment

My fifth assignment has been returned! And with it, the usual amount of feedback!

In this assignment, I failed to realise that a subgroup is normal if it is a union of conjugacy classes. So rather than form subgroups and state they were a union of conjugacy classes, I went through the process of proving each group was a sub group. Too long-winded, and frankly a waste of time. (for reference, this refers to Theorem 3.5 on properties of subgroups, p.74 in the handbook for M208).

There was another question where I had to show that a particular group had no subgroup of order 4. I took this rather literally, and found the only 4 possible combinations of elements that could be a subgroup of order 4, and then individually proved each one did not qualify either as a subgroup or as normal. Again, although I received full marks for this as it was correct, it was long-winded. It seems that all I needed to notice that a group of order 4 is either a typical cyclic group of order 4 ( $C_{4}$ ), or a Klein group ( $K_{4}$ ). A property of $C_{4}$ is that it contains an element of order 4. But the question says the group has no subgroup of order 4! Therefore, my possible subgroups must be isomorphic to $K_{4}$ . Property of $K_{4}$ : it contains three elements of order 2. This cuts the number of possibilities in half. Amazing how mindful you have to be of the properties of everything...

When having to write down subgroups of a matrix group I thought I'd written two different subgroups, but they were actually the same. One of the entries in one subgroup was $p-\frac{1}{p}$ , and the other was $\frac{1}{p}-p$ . Now, these ARE both different. But they don't constitute as different general entries in my matrix, because they are both in $\mathbb{R}-\{0\}$ . The correct answer had the second matrix with an entry that read $2\left(p-\frac{1}{p}\right)$ , which of course has a different domain: $2\mathbb{R}-\{0\}$ , so it forms a different sub group. My brain had clearly just seen two different formulas and had gone "there we go, they're different!". Not so. Domains for subgroups must be checked!

This is incredibly easy to forget: When trying to prove something is not true, don't leave it at the generalised proof that it's not true. If it's not true, provide a specific counter-example. With numbers! Remember those? The things that you count with that aren't letters? Duh! Yes, so I know to be vigilant now...

I had some trouble explaining a mapping of complex numbers in English. This is problematic. I could've programmed it and shown you the mapping and in an instant you would've thought "ah yeah!". But with lengthy explanations about how we get to the point where the real part gets mapped to the inverse of the imaginary etc etc... things can get a bit muddled. It seems I need to resort to explaining things mathematically more often, using the facts I already have.

I need to do more reading on the Isomorphism Theorem (specifically with regard to the domain of the Image that the quotient group is isomorphic to). Wow, that was a bit wordy... Ultimately I need to be more mindful about domains of Images...

So that's it! Again, happy with my high mark, but lots to remain mindful of...

Basic Groups

Group Classification

Finite Groups

…is learning mathematics.