MHB Proof that Ha=Hb: Element y in Hb Proven True for Ha

[onie mti]

i was given that if y is an element in Hb then it is an element in Ha. that is Ha=Hb

proof

I said:
since y in Hb then then y= h(1)b for h(1) in H
to show that Ha=Hb
let x be an elt in Ha
hence x=h(2)a for some h(2) in H. now i am stuck

 

[Deveno]

It's not clear what your asking here. I suspect it is:

If $y \in Hb$ AND $y \in Ha$, THEN $Ha = Hb$.

Can you confirm this?

 

[onie mti]

It's not clear what your asking here. I suspect it is:

If $y \in Hb$ AND $y \in Ha$, THEN $Ha = Hb$.

Can you confirm this?

yes I am supposed to show that every y in Hb is in Ha thefore Ha=Hb

 

[Deveno]

That's not quite the same thing.

What I said is:

$y \in Ha \cap Hb \implies Ha = Hb$

What you just said (in your last post) is:

$y \in Hb \implies y \in Ha$

which is NOT equivalent.

Your last statement is equivalent to saying $Hb \subseteq Ha$, which does not, in and of itself, force equality of the two cosets (we must also have $Ha \subseteq Hb$).

But what I said is an even STRONGER statement:

If two cosets have ANY element in common, they are the SAME coset.

*********

A typical proof runs something like this:

Suppose $y \in Ha \cap Hb$. This means that $y = ha$ for some $h \in H$, and that $y = h'b$ for some (typically different) element $h' \in H$.

From:

$ha = h'b$

we see that:

$b = h'^{-1}ha$

so that for any element $h''b \in Hb$, we have:

$h''b = (h''h'^{-1}h)a \in Ha$ since $H$ is a subgroup, and thus closed under multiplication and inversion.

This shows that $Hb \subseteq Ha$.

The proof that $Ha \subseteq Hb$ is similar, using the fact that $a = h^{-1}h'b$.

************

Another way to state these facts are:

Two (right) cosets of $H$ are either the same, or disjoint.

To see, this, note that what I have shown above is that if two cosets are not disjoint (having the common element $y$), they are equal.

 

[I like Serena]

yes I am supposed to show that every y in Hb is in Ha thefore Ha=Hb

What you write makes no sense to me.
The first part of your sentence says:

"Show that every y in Hb is in Ha."

There is no way to know this.
So this cannot be right.

It's not clear what your asking here. I suspect it is:

If $y \in Hb$ AND $y \in Ha$, THEN $Ha = Hb$.

Can you confirm this?

Erm... this makes no sense either... at least not to me.

i was given that if y is an element in Hb then it is an element in Ha. that is Ha=Hb

Let me give it a try to rephrase the original problem statement.

It is given that if y is an element in Hb then it is an element in Ha.
Show that Ha=Hb.

In symbols:
$$\forall y \in Hb: y \in Ha \implies Ha=Hb$$
Or put differently:
$$Hb \subseteq Ha \implies Ha=Hb$$

Clarify?