Orders of elements in a quotient group.

[Artusartos]

Homework Statement

I want to find the orders of the elements in $Z_8/(Z_4 \times Z_4)$, $(Z_4 \times Z_2)/(Z_2 \times Z_2)$, and $D_8/(Z_2 \times Z_2)$.

Homework Equations

The Attempt at a Solution

The elements of $Z_2 \times Z_2$ are (0,0), (1,0), (0,1), (1,1), and the elements of $Z_8$ are of course 1, 2, 3, 4, 5, 6, and 7.

So elements of $Z_8/(Z_2 \times Z_2)$ must be of the form…

$$(0,0), (1,0), (0,1), (1,1)$$
$$1+(0,0), 1+ (1,0), 1+ (0,1), 1+ (1,1)$$
$$2+(0,0), 2+(1,0),2+ (0,1), 2+(1,1)$$
$$3+(0,0), 3+(1,0), 3+(0,1), 3+(1,1)$$
…
$$7+(0,0), 7+(1,0), 7+(0,1), 7+(1,1)$$

But since the number we are adding to $Z_2 \times Z_2$ is mod 8, none of these are equal. So I don’t know how we have 2 elements $Z_2 \times Z_2$.

I’m having the same problem with [itex]D_8/(Z_2 \times Z_2)[itex].<br /> We know that $D_8 = \{e, b, b^2, b^3, a, ba, b^2a, b^3a \}$<br /> <br /> $$(0,0), (1,0), (0,1), (1,1)$$<br /> $$b+(0,0), b+(1,0), b+ (0,1), b+(1,1)$$<br /> …<br /> $$b^3a+(0,0), b^3a+(1,0), b^3a+(0,1), b^3a+(1,1)$$<br /> <br /> But, for example, we know that $b$ has order 4, right?[/itex][/itex]

 

[Dick]

Homework Statement

I want to find the orders of the elements in $Z_8/(Z_4 \times Z_4)$, $(Z_4 \times Z_2)/(Z_2 \times Z_2)$, and $D_8/(Z_2 \times Z_2)$.

Homework Equations

The Attempt at a Solution

The elements of $Z_2 \times Z_2$ are (0,0), (1,0), (0,1), (1,1), and the elements of $Z_8$ are of course 1, 2, 3, 4, 5, 6, and 7.

So elements of $Z_8/(Z_2 \times Z_2)$ must be of the form…

$$(0,0), (1,0), (0,1), (1,1)$$
$$1+(0,0), 1+ (1,0), 1+ (0,1), 1+ (1,1)$$
$$2+(0,0), 2+(1,0),2+ (0,1), 2+(1,1)$$
$$3+(0,0), 3+(1,0), 3+(0,1), 3+(1,1)$$
…
$$7+(0,0), 7+(1,0), 7+(0,1), 7+(1,1)$$

But since the number we are adding to $Z_2 \times Z_2$ is mod 8, none of these are equal. So I don’t know how we have 2 elements $Z_2 \times Z_2$.

I’m having the same problem with [itex]D_8/(Z_2 \times Z_2)[itex].<br /> We know that $D_8 = \{e, b, b^2, b^3, a, ba, b^2a, b^3a \}$<br /> <br /> $$(0,0), (1,0), (0,1), (1,1)$$<br /> $$b+(0,0), b+(1,0), b+ (0,1), b+(1,1)$$<br /> …<br /> $$b^3a+(0,0), b^3a+(1,0), b^3a+(0,1), b^3a+(1,1)$$<br /> <br /> But, for example, we know that $b$ has order 4, right?[/itex][/itex]

[itex][itex] <br /> Your first question doesn't make sense to me whatsoever. The quotient G/H is defined when H is a subgroup of G. $Z_2 \times Z_2$ is not a subgroup of $Z_8$. It doesn't even have a subgroup isomorphic to $Z_2 \times Z_2$. $D_8$ does so I think you can make some sense of that.[/itex][/itex]

 

[Artusartos]

Your first question doesn't make sense to me whatsoever. The quotient G/H is defined when H is a subgroup of G. $Z_2 \times Z_2$ is not a subgroup of $Z_8$. It doesn't even have a subgroup isomorphic to $Z_2 \times Z_2$. $D_8$ does so I think you can make some sense of that.

Thanks a lot.