Finding Cosets of Subgroups in Groups

[roam]

Homework Statement

[PLAIN]http://img641.imageshack.us/img641/8448/63794724.gif

The Attempt at a Solution

Firstly, how do I list the elements of H?

According to Lagrange's theorem if H is a subgroup of G then the number of distinct left (right) cosets of H in G is |G|\|H|.

So I must find the orders of G and H:

Since $$U(5)={1,2,3,4})$$ and $$\mathbb{Z}_4 = \{ 1,2,3,4 \})$$, the order of

$$G=U(5) \oplus \mathbb{Z}_4 = (1,1),(1,2),(1,3),(1,4), (2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)$$

So G has order 16.

H is generated by the element (4,3), where 4 is an element of U(5) and 3 is from Z₄. I know that $$| \left\langle (4,3) \right\rangle | = |(4,3)|$$. So I think

|H|=|(4,3)|=lcm(|4|,|3|)=12

Going back to lagrange's theorem |G|\|H|=16\12=4\3

But how could the number of cosets be a fraction? Could anyone please show me what I did wrong?

 

[ystael]

Three notational problems:

1. What is $$U(5)$$? Is it the group $$(\mathbb{Z}/(5))^\times$$, the multiplicative group of units of the integers modulo 5?

2. If $$\mathbb{Z}_4$$ represents the additive group of integers modulo 4, it is conventional to choose representatives $$\{0, 1, 2, 3\}$$ rather than $$\{1, 2, 3, 4\}$$.

3. The order of an element of a group (as opposed to the order of a group) is not written with absolute value bars. Some people write $$o(g)$$ for the order of an element $$g$$, which as you know equals the order $$|\langle g\rangle|$$ of the subgroup it generates. I think this notational confusion caused your mistakes below.

Now, supposing I've guessed correctly about your notational issues, you have made two mistakes: the order of $$4$$ in $$U(5) = (\mathbb{Z}/(5))^\times$$ is not $$4$$, and the order of $$3$$ in $$\mathbb{Z}_4$$ is not $$3$$. Figure out what the correct orders are, and that should solve your problem.

You also haven't listed the elements of $$H = \langle (4, 3) \rangle$$, which the question asked for.