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Let $H$ be a subgroup of group $G$, and let $a$ and $b$ belong to $G$. Then, it is known that

$$ aH=bH\qquad\text{or}\qquad aH\cap bH=\emptyset $$

In other words, $aH\neq bH$ implies $aH\cap bH=\emptyset$. What can we say about the statement

"If $aH\neq Hb$, then $aH\cap Hb=\emptyset$" ?

[EDITED:] What I think is that when $G$ is Abelian, this can be true since $aH=Ha$ for any $a\in G$. But what if $G$ is non-Abelian? How should I go on?

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Try $H$ being a normal subgroup. –  KCd Oct 10 '11 at 3:38
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Note that $H$ being abelian does not imply $aH = Ha$ "for any" $a$ in $G$. If $a$ is not in $H$, then... –  KCd Oct 10 '11 at 3:39
    
@KCd: Oops, corrected. Thanks. –  Jack Oct 10 '11 at 3:41

2 Answers 2

up vote 8 down vote accepted

The condition $aH=Ha$ characterizes a special kind of subgroup, called a normal subgroup. They play a very important role.

Theorem. Let $G$ be a group, and let $H$ be a subgroup of $G$. The following are equivalent:

  1. Every left coset of $H$ is also a right coset of $H$.
  2. $aH = Ha$ for every $a\in G$.
  3. $aHa^{-1} = H$ for every $a\in G$.
  4. $aHa^{-1}\subseteq H$ for every $a\in G$.
  5. The equivalence relations $a\equiv_H b$ and $a {}_H\equiv b$ are the same equivalence relation.
  6. There exists a group $K$ and a homomorphism $f\colon G\to K$ such that $H=\mathrm{ker}(f)$.

(Remember that $a\equiv_H b$ if and only if $ab^{-1}\in H$; and $a{}_H\equiv b$ if and only if $a^{-1}b\in H$).

If $H$ is a normal subgroup, then your implication holds, exactly the same way as it holds for abelian groups.

Conversely, suppose that $H$ is not a normal subgroup. Then there exists an $a$ such that $aH\neq Ha$; but $a\in aH\cap Ha$. Thus, $aH\neq Ha$ but $aH\cap Ha\neq\emptyset$, so the implication does not hold. That is: your desired implication is equivalent to $H$ being a normal subgroup of $G$.

So we could add a seventh point to the theorem: "if $aH\cap Hb\neq\emptyset$, then $aH=Hb$."

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Thanks a lot for your work. I don't understand your notations in the statement No. 5. I haven't seen that in the textbook before. Does $\equiv_H$ mean equivalence relation with respect to "left cosets"? And is No.6 supposed to be "...$H=\ker(f)$"? –  Jack Oct 10 '11 at 3:54
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@Jack: Yes on the second. As to the first, I gave the definition right after 6. $\equiv_H$ is called "congruent on the right modulo $H$", and ${}_H\equiv$ is "congruent on the left modulo $H$". They are meant to be a generalization of congruence for integers (where $a\equiv b\pmod{m}$ if and only if $a-b\in m\mathbb{Z}$). It gives another way of defining cosets, since the left cosets are precisely the equivalence classes relative to ${}_H\equiv$, and the right cosets are precisely the equivalence classes relative to $\equiv_H$. –  Arturo Magidin Oct 10 '11 at 4:03
    
Fari enough. Thanks. –  Jack Oct 10 '11 at 13:12

It is sometimes true and sometimes false.

For example, if $H$ is a normal subgroup of $G$, then it is true.

If $H$ is the subgroup generated by the permutation $(12)$ inside $G=S_3$, the symmetric group of degree $3$, then $(123)H\neq H(132)$, yet $(13)\in(123)H\cap H(132)$

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+1 for the simplicity. –  Jack Oct 10 '11 at 3:55
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@Jack In fact, checking the smallest non-abelian group should have been the first thing for you to do before posting here. –  Alex B. Oct 10 '11 at 4:16

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