# Questions Involving Cosets

• Mar 14th 2009, 03:12 PM
Janu42
Questions Involving Cosets
1. Let G = Q8 (quaternion group). Find the right cosets of H in G for:
a) H = <J>
b) H = <-I>

2. Find the right cosets of the subgroup H = <(1,1)> in Z2 x Z4

3. Let H be a subgroup of a group G and define H congruent on G by letting x congruent to y IFF x^-1y is in H.
a) Show that H congruent is an equivalence relation on G
b) Show that the equivalence classes under H congruent are the left cosets of H in G.
c) Show that for a,b in G, aH = bH iff a^-1b is in H.

4. Let G = Q8 (quaternion group). Find [G : H] for H = <-I>, H = <K>, and H = <-L>.

5. a) In G = (Z48, +), find [G : H] for H = <32>.
(+ is actually addition mod 48, if it was unclear). There are two more parts to this question but if someone could tell me how to do this one, I can follow with the others.
• Mar 14th 2009, 05:54 PM
ThePerfectHacker
Quote:

Originally Posted by Janu42
1. Let G = Q8 (quaternion group). Find the right cosets of H in G for:
a) H = <J>

We have $\displaystyle H = \{1,-1,j,-j\}$ then $\displaystyle iH=\{i,-i,k,-k\}$.
Thus, these two sets are the cosets.

Quote:

2. Find the right cosets of the subgroup H = <(1,1)> in Z2 x Z4
First $\displaystyle H=\{ (0,0),(1,1),(0,2),(1,3)\}$ thus $\displaystyle (0,1)+H = \{(0,1),(1,2),(0,3),(1,0)\}$.
These are the cosets.

Quote:

3. Let H be a subgroup of a group G and define H congruent on G by letting x congruent to y IFF x^-1y is in H.
a) Show that H congruent is an equivalence relation on G
b) Show that the equivalence classes under H congruent are the left cosets of H in G.
c) Show that for a,b in G, aH = bH iff a^-1b is in H.
Of course, $\displaystyle x\equiv x$ since $\displaystyle x^{-1}x = e\in H$ thus it is reflexsive. It is symmetric because if $\displaystyle x\equiv y\implies x^{-1}y\in H$ but then $\displaystyle (x^{-1}y)^{-1}\in H\implies y^{-1}x\in H\implies y\equiv x$. Finally, it is transitive because if $\displaystyle x\equiv y\text{ and }y\equiv z$ this means $\displaystyle x^{-1}y,y^{-1}z\in H$ but then $\displaystyle (x^{-1}y)(y^{-1}z) = x^{-1}z\in H\implies x\equiv z$. This completes the proof that $\displaystyle \equiv$ is an equivalence relation.

Now we wish to prove $\displaystyle [a] = aH$ where $\displaystyle [a] = \{ x\in H | a\equiv x\}$ and $\displaystyle aH = \{ ah|h\in H\}$. Remember to show two sets are equal show that any element in $\displaystyle [a]$ is an element in $\displaystyle aH$ and any element in $\displaystyle aH$ is in $\displaystyle [a]$. If $\displaystyle x\in [a] \implies a\equiv x \implies a^{-1}x\in H$ and so $\displaystyle a^{-1}x = h$ for some $\displaystyle h\in H$, but then this means $\displaystyle x=ah\in aH$. The reserve direction is straightforward too.

Remember that $\displaystyle aH = [a]\text{ and }bH = [b]$ so for $\displaystyle [a] = [b]$ we require that $\displaystyle a\equiv b$, and so $\displaystyle a^{-1}b\in H$.

Quote:

4. Let G = Q8 (quaternion group). Find [G : H] for H = <-I>, H = <K>, and H = <-L>.

5. a) In G = (Z48, +), find [G : H] for H = <32>.
(+ is actually addition mod 48, if it was unclear). There are two more parts to this question but if someone could tell me how to do this one, I can follow with the others.
Here is a hint for these problems. If $\displaystyle G$ is a finite group then $\displaystyle [G:H] = \tfrac{|G|}{|H|}$.

This is Mine 12,2:):)th Post!!!
• Mar 14th 2009, 07:51 PM
Janu42
Sorry, I don't understand the answer to #1. You put H and then iJ? I'm sorry if I'm missing something here, but I'm confused...
• Mar 14th 2009, 07:52 PM
ThePerfectHacker
Quote:

Originally Posted by Janu42
Sorry, I don't understand the answer to #1. You put H and then iJ? I'm sorry if I'm missing something here, but I'm confused...

Sorry, I meant iH not iJ. It is fixed now.