# Abstract Algebra

1. Mar 22, 2005

### Oxymoron

Question 1
Determine the quotient group $$(\mathbb{Z}_2\times\mathbb{Z}_4)/\langle(1,2)\rangle$$

$$\langle(1,2)\rangle$$ is a cyclic subgroup $$H$$ of $$\mathbb{Z}_2\times\mathbb{Z}_4$$ generated by $$(1,2)$$. Thus

$$H=\{(0,0),(1,2)\}$$

Since $$\mathbb{Z}_2\times\mathbb{Z}_4$$ has 2.4 = 8 elements, and $$H$$ has 2 elements, all cosets of $$H$$ must have 2 elements, and $$(\mathbb{Z}_2\times\mathbb{Z}_4)/H$$ must have order 4.

Possible abelian groups of order 4 are

$$\mathbb{Z}_2\times\mathbb{Z}_2$$
$$\mathbb{Z}\times\mathbb{Z}_4$$

But I dont know how to work out which one is isomorphic to $$(\mathbb{Z}_2\times\mathbb{Z}_4)/\langle(1,2)\rangle$$

2. Mar 22, 2005

### Data

Which of $$\mathbb{Z}_2 \times \mathbb{Z}_2$$, $$\mathbb{Z}_1 \times \mathbb{Z}_4$$, and $$(\mathbb{Z}_2 \times \mathbb{Z}_4)/H$$ are cyclic?

3. Mar 22, 2005

### Oxymoron

I think $$\mathbb{Z}\times\mathbb{Z}_4$$ is cyclic because it is generated by $$a^n$$ for all $$n=0,1,\dots$$ where $$a=(1,2)$$ and $$n = 2$$

4. Mar 22, 2005

### Data

Well, as I posted, you really mean $\mathbb{Z}_1 \times \mathbb{Z}_4$ (or equivalently just $\mathbb{Z}_4$), since $\mathbb{Z} \times \mathbb{Z}_4$ doesn't have order 4 at all (not even finite order!).

But it is indeed cyclic (generated by $(0, 1)$ or just $1$ if you choose the $\mathbb{Z}_4$ variety~).

Is $(\mathbb{Z}_2 \times \mathbb{Z}_4)/H$ cyclic?

Last edited: Mar 22, 2005
5. Mar 22, 2005

### Oxymoron

Yes, it is cyclic. Isn't there a theorem that says a quotient group of cyclic group is cyclic?

So wouldn't $$\mathbb{Z}\times\mathbb{Z}_4$$ be the answer? Or am I missing something?

6. Mar 22, 2005

### Data

I don't remember such a theorem. Let's just write down the elements of $(\mathbb{Z}_2 \times \mathbb{Z}_4)/H[/tex]. This set is just $$\{ (1, 0) + H, \; (1, 1) + H, \; (0, 1) + H, \; (0, 0) + H \}$$. Does it have a generator? Remember the group is abelian and [itex] H + H = H$.

Edit: Actually, there may be. I really don't remember! Anyways, in this case the group in question is indeed cyclic (and thus isomorphic to $\mathbb{Z}_4$).

Last edited: Mar 22, 2005
7. Mar 22, 2005

### Oxymoron

$$\mathbb{Z}_2\times\mathbb{Z}_4 = \{(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)\}$$

But this can be collapsed into

$$\mathbb{Z}_4 = \{(0,0),(0,1),(0,2),(0,3)\} = \mathbb{Z}_2\times\mathbb{Z}_2$$

So then

$$(\mathbb{Z}_2\times\mathbb{Z}_4)/H = \{(0,0)+H,(0,1)+H,(0,2)+H,(0,3)+H\}$$

where $$H=\langle(1,2)\rangle$$

and its generator is $$(1,2)$$

Last edited: Mar 22, 2005
8. Mar 22, 2005

### Data

I think the theorem you're thinking of is that if $G$ is cyclic then every quotient group $G/H$ is cyclic. In our case we can't use that (since $H$ is cyclic, not $\mathbb{Z}_2 \times \mathbb{Z}_4$). There's no theorem the way we need it

Last edited: Mar 22, 2005
9. Mar 22, 2005

### Data

I wouldn't include the line saying "can be collapsed into" (what does that mean?). And I think in your second TeX line (the one I think you should remove, because the first equality isn't true~) you meant $= \mathbb{Z}_1 \times \mathbb{Z}_4$, not $= \mathbb{Z}_2 \times \mathbb{Z}_2$!

The rest looks good though~

Last edited: Mar 22, 2005
10. Mar 22, 2005

### Oxymoron

Ok, I fixed it up a bit.

Let me get this straight. The quotient group under investigation is
$$(\mathbb{Z}_2\times\mathbb{Z}_4)/\langle(1,2)\rangle$$.

Here $$\langle(1,2)\rangle$$ is the cyclic subgroup $$H$$ of $$\mathbb{Z}_2\times\mathbb{Z}_4$$ generated by $$(1,2)$$. Thus

$$H = \{(0,0),(1,2)\}$$

Since $$\mathbb{Z}_2\times\math{Z}_4$$ has 8 elements and $$H$$ has 2 elements, all cosets of $$H$$ must have 4 elements, and $$(\mathbb{Z}_2\times\mathbb{Z}_4)/H$$.

In additive notation, the cosets are

$$H = \{(0,0)+H,(0,1)+H,(0,2)+H,(0,3)+H\}$$

Since we can compute by choosing representatives $$(0,0),(0,1),(0,2),(0,3)$$ it is clear that $$(\mathbb{Z}_2\times\mathbb{Z}_4)/H$$ is isomorphic to $$\mathbb{Z}_4$$

Note that this is what we would expect, since in a factor modulo $$H$$, everything in $$H$$ becomes the identity element; that is we are essentially setting everthing in $$H$$ to zero. Thus the whole factor $$\mathbb{Z}_2$$ of $$\mathbb{Z}_2\times\mathbb{Z}_4$$ is collapsed, leaving just the second factor $$\mathbb{Z}_4$$.

11. Mar 22, 2005

### Data

looks good

12. Mar 22, 2005

### Oxymoron

Question 2

Show that ifa group $$G$$ has exactly one subgroup $$H$$ of given order, then this subgroup must be normal.

I figured that since every group has at least two subgroups: the whole group itself (the improper subgroup) $$G$$ and the trivial subgroup $$\{e\}$$.

If a group $$G$$ has exactly one subgroup then $$G = \{e\}$$ and the group in question is actually the trivial group. This is the only way that I can see that a group can have exactly one subgroup - if it is the trivial group.

So if $$G$$ is the trivial group, then it consists of only one element, namely the identity. The group consisting of one element, $$e$$ is commutative. ie $$e*a=a*e$$ where $$a \in G$$. Obviously $$a=e$$ if $$G$$ has only one element.

Hence $$G$$ is abelian. And all subgroups of abelian groups are normal.

Proof

A subgroup is normal if its left and right cosets coincide. That is, $$gH=Hg \, \forall \, g \in G$$. Since $$g=e$$ then we have $$eH=He$$ which is trivial because $$G$$ is abelian.

Therefore the subgroup $$H$$ of $$G$$ is normal.

13. Mar 22, 2005

### Data

You are misinterpreting the question (at least the way I read it). It is not saying that $G$ has only one subgroup. It says that $G$ has only one subgroup $H$ of a particular order, say $n$, ie. if $J \leq G$ and $|J| = n$ then $J = H$.

14. Mar 22, 2005

### Oxymoron

Ok, lets assume that you read the question correctly (which is likely). Then I have to show that the subgroup is normal given that it is the only subgroup with a particular order $$n$$.

This means that $$G$$ has only one subgroup with $$n$$ elements, all other subgroups have some other order.

For a subgroup of a group to be normal, then the following are equivalent
$$ghg^{-1} \in H \, \forall \, g\in G \mbox{and} h\in H$$
$$gHg^{-1} = H \, \forall \, g\in G$$
$$gH = Hg \, \forall g\in G$$

But I don't know how to start. That is, none of the conditions for a subgroup to be normal, involve the order of the subgroup.

Any hints?

Last edited: Mar 22, 2005
15. Mar 23, 2005

### fourier jr

also
N normal <==> NaNb = Nab for a, b not in subgroup N

16. Mar 24, 2005

### Data

Sorry I haven't replied in a while.

Here's a hint:

Recall that $H$ normal in $G$ if and only if $gHg^{-1} = H \mbox{ for every } g \ \mbox{in} \ G$. Assume $H$ isn't normal. Then there is some $g$ in $G$ such that $gHg^{-1} \neq H$. For this $g$, consider the set $gHg^{-1}$ (Hint: Try to prove that it is a subgroup of $G$ of the same order as $H$, to get a contradiction).

Last edited: Mar 24, 2005
17. Mar 24, 2005

### Data

Figured I'd put this back at the top in case you still need it, since you're online right now~

18. Mar 24, 2005

### Oxymoron

The condition for $$H$$ being normal is

$$ghg^{-1} \in H\quad \forall g \in G,\, \forall h \in H$$

Assume that $$H$$ isn't normal (proof by contradiction)

Then $$\exists \, k \in G$$ such that

$$khk^{-1} \notin H \quad \forall h \in H$$

Now consider the set

$$A = \{khk^{-1} | \, \forall h \in H\}$$

We now show that this is a subgroup of $$G$$ and that it has the same order as $$H$$.

For $$h_1,h_2 \in H, \, \exists\, kh_1k^{-1},\, kh_2k^{-1} \in A$$ by definition. Now,

$$(kh_1k^{-1})(kh_2k^{-1}) = (kh_1)(k^{-1}k)(h_2k^{-1})$$
$$= kh_1(e)h_2k^{-1}$$
$$= k(h_1h_2)k^{-1}$$

But $$h_1,h_2 \in H$$ as $$H$$ is a subgroup by assumption.

Hence $$kh_1,h_2k^{-1} \in A$$ which implies that $$A$$ is a subgroup of $$G$$

Now so far we have been using the fact that $$H$$ is not normal. And from this we have got a subgroup $$A$$. But it is obvious that for each $$h \in H$$ we can make an $$a \in A$$ by

$$kh_1k^{-1} = kh_2k^{-1} \Leftrightarrow h_1 = h_2$$

the left and right cancellation laws. Hence $$H$$ and $$A$$ have the same order. Alternatively I think I could have proven this by setting a bijection. Anyway, $$H$$ and $$A$$ having the same order is a contradiction since $$H$$ is the only subgroup with a given order. Hence $$H$$ MUST be normal.

Last edited: Mar 24, 2005
19. Mar 24, 2005

### Data

Everything looks great, except for one detail:

You cannot assume $\exists k \in G$ such that

$$khk^{-1} \notin H \; \forall h \in H$$.

This is too strong.

All you can assume is

$\exists k \in G$ such that

$$\{ khk^{-1} | h \in H \} \neq \{ h \in H\}$$

there is a subtle difference, but it does not affect the rest of the proof.

Also, you must show that $e \in A$, but this is trivial. Also, that $x \in A \Longrightarrow x^{-1} \in A$ (also easy).

You are quite correct that the trivial bijection (and, in fact, isomorphism)

$$f : H \longrightarrow A$$
$$\forall \ h \in H, \ f(h) = khk^{-1}$$

would also have worked nicely

Last edited: Mar 25, 2005
20. Mar 25, 2005

### Oxymoron

Thanks Data for your input. I'd have to say, you have a knack for mentoring.

Here is another question I have been working on.

Let $$\textsc{Q}$$ be the subgroup of $$GL(2,\mathbb{C})$$ generated by

$$\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array}\right) \quad \mbox{and} \quad \left(\begin{array}{cc} 0 & i \\ i & 0 \\ \end{array}\right)$$

1. Show that $$\textsc{Q}$$ is a nonabelian group of order 8.
2. Is $$\textsc{Q}$$ isomorphic to the dihedral group $$\mathcal{D}_4$$?