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$C_4 = \{e, a, a^2, a^3\}$

A normal subgroup of $C_4$ is $C_2 = \{e, a^2\}$

So I am wondering what the quotient group $G/N$ looks like in this case.

Ie. where $G = C_4$ and $N = C_2$.

The right (or left) cosets of $N$ are

$Ne = \{e, a^2\}$ and

$Na = \{a, a^3\}$

$G/N$ is the group formed by these cosets so I have it as = $\{ \{e, a^2\}, \{a, a^3\} \}$

Is that right...it seems weird having each element being a set of elements..?

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2 Answers

up vote 2 down vote accepted

Yes, the elements of $G/N$ are cosets, which are themselves subsets of $G$. But you should think of each individual coset as a single element of the set $G/N$. But $G/N$ is not just a set -- it's a group.

The operation is multiplication of cosets in the following way: $$(g_1 N) \cdot (g_2 N) = g_1 g_2 N.$$ In other words, you multiply two cosets as follows. You take a representative ($g_1, g_2$) of each coset, multiply the two representatives (in the given order, getting $g_1 g_2$) and then the product $(g_1 N) \cdot (g_2 N)$ is the coset containing $g_1 g_2$, namely $g_1 g_2 N$. However, it turns out that this product is well-defined (you don't get different answers for different choices of representatives from your coset) only if the subgroup $N$ is normal.

In your example, you can make a Cayley table with the rows and columns given by your cosets. You should find that the multiplication table looks very familiar -- $G/N$ in your example is isomorphic to the only group of order $2$ there is (up to isomorphism).

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Is it possible to multiply them element-wise. Like say I have from my example above. $Na = \{a, a^3\}$ and I want $NaNa$. Rather than saying $NaNa = N(aa) = Na^2 = N$, is it possible to have $NaNa = \{a, a^3\}\{a, a^3\} = .... $?? I don't see how you could multiply two sets like that, it seems you can only do it using the $NaNb = Nab$ formula...is that correct? –  sonicboom Nov 12 '12 at 21:32
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A consequence of the product being well defined is that no matter which pair of representatives you choose, you end up with an element of the correct coset. Therefore, if you multiply every element in the first set with every element of the second set, at the very least you get an element of the correct coset. With a little more work, you can see that you'll get every element of the correct coset. So yes, you can multiply all the elements by all the elements to get the set of all products, and it will be the coset you're looking for. –  Mark S. Nov 12 '12 at 21:37
    
Thanks, I see now it works alright. –  sonicboom Nov 12 '12 at 23:42
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That's right. As long as you know what the whole cosets are, you can write them as $[e]$ and $[a]$, where $[\cdot]$ means "the equivalence class of $\cdot$". Also, now that you know $G/N$ has two elements, you should be able to figure out which named group(s) $G/N$ is isomorphic to.

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