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Problem 84 (Properties Of Cosets Part Two)

Let $H$ be a subgroup of $G$. Let $a,b\in G$. Prove the following facts about cosets.

  1. We have $Ha=Hb$ if and only if $ab^{-1}\in H$.
  2. We must have $Ha=Hb$ or $Ha\cap Hb=\emptyset$.
  3. We have $Ha=aH$ if and only if $aHa^{-1}=H$.

We've already seen all of these properties before when studying modular arithmetic. Let $G=\mathbb{Z}$ and let $H=n\mathbb{Z}$ for some $n\in\mathbb{N}$. The cosets of $H$ are $r+n\mathbb{Z}$ for $0\leq r<n$.

  • The first property above states that $a+n\mathbb{Z}=b+n\mathbb{Z}$ if and only if $a-b\in n\mathbb{Z}$. Wait, this just says two numbers have the same remainder if and only if their difference is a multiple of $n$.
  • The second property above states that either $a$ and $b$ have the same remainder, or they don't. It basically states that remainders are unique.
  • The third property is trivial for $\mathbb{Z}$ because $\mathbb{Z}$ is an Abelian group and we always know $a+n\mathbb{Z}=n\mathbb{Z}+a$.

1. We will show $Ha = Hb$ if and only if $ab^{-1} \in H$.

Suppose that $ab^{-1} \in H$. Since we know that $ab^{-1}b = a$, we see that $a\in Hb$. We know that $b\in Ha$ if and only if $Hb = Ha$, so this shows that $Ha=Hb$.

For the other direction, suppose that $Ha=Hb$. Then $a\in Hb$. Since the set product is associative, we know $(Hb)b^{-1} = H(bb^{-1}) = He = H$. Since we know $a\in Hb$, this means that $ab^{-1}\in H$.

2. We will show that we have $Ha = Hb \vee Ha\cap Hb = \emptyset$.

Suppose that $Ha\cap Hb \neq \emptyset$. We will show that $Ha = Hb.$ Let $n\in Ha\cap Hb.$ Then choose $x\in H$ such that $xa = h$ and $y\in H$ such that $yb=h$. This means that $x=ha^{-1}$ and $y=hb^{-1}$. We know that $x^{-1} = ah^{-1}$. This means that $x^{-1}y = ah^{-1}hb^{-1}=aeb^{-1} = ab^{-1}$. We showed in part one that this is sufficient to show that $Ha=Hb$.

3. Last, we will show that $Ha = aH$ if and only if $aHa^{-1} = H$.

Suppose that $aHa^{-1} = H$. Then we see that $aHa^{-1}a = Ha = aHe = aH$ when we multiply on the right by $a$.

For the other direction, suppose that $Ha = aH$. Then we see that $Haa^{-1} = aHa^{-1} = H$ when we multiply on the right by $a^{-1}$.

From Ben: You have not fixed the problem with this section. The takeaways (punchlines) of your computations are still not coming accross, and your work appears circular. Come by.
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