Let H be a subgroup of a group G. Prove that the following statements are equivalent.
(a) For all $a,b \in G, (aH)(bH)$ is a left coset of $H$ in $G$.
(b) For all $a,b \in G, (aH)(bH) = (ab)H$.
(c) For all $a \in G, (aH)(a^{-1}H) = H.$
(d) For all $a \in G, aHa^{-1} \subseteq H.$
(e) For all $a\in G, aHa^{-1} = H.$
(f) For all $a\in G, aH = Ha.$
My attempt:
(a) implies (b):
I can show $abH \subseteq aHbH$ since $abh_1 = a1_Hbh_1$ but i can't show the other direction.
NEED HELP
(b) implies (c) :
Let $b = a^{-1}$
(c) implies (d):
For all $a \in G$, $ah_1a^{-1}h_2 = h_3$ and hence $ah_1a^{-1} = h_3(h_2)^{-1} \in H$. Therefore, $aHa^{-1} \subseteq H.$
(d) implies (e):
We have to show that all elements H can be expressed as $ah'a^{-1}$. For all $x \in H$, $ x= a(a^{-1}xa)a^{-1}$. $(a^{-1}xa) \in H$ from (d) by letting $a = a^{-1}$. Hence , $H \subseteq aHa^{-1}$. Together with (d), $aHa^{-1} = H$.
(e) implies (f):
$aHa^{-1} = H$ hence $aH = Ha$ by multiplying a on the right.
(f) implies (a):
NEED HELP.
Are my proof OK?
Best Answer
(a) implies (b): You know that $(aH)(bH)$ is a left coset of $H$ in $G$, and it intersects non-trivial the left coset $abH$. If two left cosets intersect each other, they coincide, so you must have equality.
(f) implies (a): I'll show that $(aH)(bH)$ is the left coset $abH$. Because $H$ is a subgroup of $G$, the set of all products of two elements in $H$, namely $HH$, is equal to $H$. Thus:
$$ (aH)(bH)\overset{(f)}=(aH)(Hb)=a(HH)b=aHb=a(Hb)\overset{(f)}=a(bH)=abH $$
Edit: A sketch on how to prove that distinct cosets do not intersect. Let $aH,bH$ be two cosets, and suppose that they are distinct. Suppose (towards a contradiction) that they do intersect at some point $c\in aH \cap bH$.
By definition, we find some $h,k\in H$ such that $c=ah=bk$. Thus $b=ahk^{-1}$. Now let $x=bh^{\prime}$ be any element of $bH$. Then: $$ x=bh^{\prime}=ahk^{-1}h^{\prime} $$ and $hk^{-1}h^{\prime}$ is in $H$ as $H$ is a subgroup. Thus $x\in aH$, and we have the inclusion $bH \subseteq aH$. Similarly, one has that $a=bkh^{-1}$ and repeats the argument to prove that $aH \subseteq bH$