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I'm still a little confused about cosets. Maybe someone could look my proof over, please?

  1. If $G$ is a group and $H$ is a subgroup, $a, b \in G$, need to prove $Ha = H$, iff $a \in H$.

(->) Let $Ha = H$. Suppose $s \in Ha$. Then $s = ha$ for some $h \in H$. Since $a \in H, h \in H$, then $s \in H$. So $Ha \subset H$.

(<-) Let $t \in H$. Then $$ t = t e = t (aa^{-1}) = t(a^{-1}a) = (ta^{-1})a$$ Since $a \in H$, then $a^{-1} \in H$, and $t \in H$, so $ta^{-1} \in H$. Therefore, $t \in Ha$ and $H \subset Ha$.

  1. How does one go about listing elements of the coset of H, eg. $G = S_3$ and $H = \{\epsilon, \beta, \delta\}$?
Bob
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2 Answers2

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Your proof really only shows that $a\in H$ implies that $Ha=H$. That proof seems fine to me. Your -> and <- steps prove the two inclusions directions. You do need to prove the other direction as well: suppose that $Ha=H$ and show that $a\in H$. (This direction is actually pretty easy - just choose the right element of $H$ to multiply by $a$.)

For your second question, in this particular case, since $S_3$ has six elements and $H$ has three, you know that there is only one other coset. So take any element $a\in G\backslash H$ and multiply it by each of the elements of $H$. This gives you $Ha$, which you know cannot be equal to $H$ by the first part, so it must be the other coset.

rogerl
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Note that $a=ea\in Ha$ so that $Ha=H$ immediately leads to $a\in H$.

Your proof of the converse is okay, except that the part $\quad\cdots=t(aa^{-1})\cdots\quad$ is irrelevant and confusing. It can just be left out.

drhab
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