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I know how to prove that it happens, by proving that the left coset definition actually is an equivalence relation. Then, it's proved that it partitions the set, since equivalence relations do it.

However, this exercise asks me to prove it in a diferente way. First it asks me to show that:

a) The union of the left cosets is equal to $G$

b) If $gH \cap g'H \neq \emptyset$ then $gH = g'H$

for $a)$ I'm thinking about the following:

$gH = \{gh, h\in H\}$

So if $G = \{g_1, g_2, \cdots g_n\}$

We would have the following left cosets:

$$g_1H = \{g_1h, h\in H\}$$ $$g_2H = \{g_2h, h\in H\}$$ $$\cdots$$ $$g_nH = \{g_nh, h\in H\}$$

The union of all these sets will include all the $g's$, since for each set

$$g_k = \{g_kh, h\in H\}$$

we have

$$ge \in g_k = \{g_kh, h\in H\}$$

where $e$ is the identity.

Then if we make the union of all these sets we'll have at least all the elements of $g$. The other elements are merely $gh$ for some $h$. But since $gh\in G$ they would be repeated elements in the union, doesn't matter. So, the union of all left cosets of $H$ in $G$ is $G$.

Is my reasoning correct?

Also, what could I do to prove $b)$?

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  • $\begingroup$ But (1)-(2) are exactly what an equivalence class does to a set: (1) it partitions it and thus the union of the equiv. classes is the whole set, and (2) the equiv. classes are pairwise disjoint...if as you say you can prove this with equivalence classes, what is then to prove here?? $\endgroup$
    – Timbuc
    Jul 11, 2015 at 7:27
  • $\begingroup$ @Timbuc when I proved that it is equiv. class, I proved the 3 conditions necessary, and then used the theorem that it partitions. I needed a proof for this case. $\endgroup$ Jul 11, 2015 at 7:29

2 Answers 2

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Here is the proof for (b)

suppose that there is an element $c \in gH \cap g^\prime H$ so that means that $c \in gH$ and $c \in g^\prime H$ then this means that there are elements $a,b \in H$ such that $c = ga$ and $c = g^\prime b$

now this implies that $$c=ga=g^\prime b$$ which also implies that $$g = g^\prime b a^{-1}$$ because we can right multiply by $a^{-1}$ on each side.

However, observe that $ba^{-1} \in H$ (Closure of groups)

and so $$g \in g^\prime H$$

and so for any $a \in H$, $ga \in g ^\prime H$

which implies that $$gH \subseteq g^\prime H$$ By a symmetrical argument ,$$g^\prime H \subseteq gH$$

and hence $$gH = g^\prime H$$

Your argument for part (a) Looks decent :)

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Part (1) looks perfectly fine.
For part (2), assume that $g_1H\cap g_2 H\ne(e).$

Then it follows that, for some $h_1,h_2\in H, g_1h_1=g_2h_2$. Then $g_1h_1h_2^{-1}=g_2$. But since $h_1h_2^{-1}\in H$, it follows that $g_2\in g_1H$. So $g_2=g_1h_3$ for some $h_3\in H$. Thus $h_3=g_1^{-1}g_2$. And so, using the closure of $H$, all elements of the form $g_1^{-1}g_2h$ are contained in $H$, and so all elements of the form $g_1g_1^{-1}g_2h=g_2h$ are contained in $g_1H$. And so you can conclude that $g_2H\subseteq g_1H$. Using similar logic gives you that $g_1H\subset g_2H$, and hence, $g_1H=g_2H$.

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