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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..?

2 Answers 2

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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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    Thanks, I see now it works alright.2012-11-12
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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.