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Assume $G$ is a group extension of $Z$ by $H$ (i.e $Z$ is a normal subgroup of $G$). I have trouble proving that the elements of $G$ can be uniquely represented as $(z,h)$, $z \in Z$, $h \in H$.

What I used so far is a preimage $x$ of each $h \in H$, to form a coset $xZ$. Since the cosets of $G$ form a partition of $G$, this should give me the result. However I'm unsure I'm getting all the cosets by this construction.

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If it is not a split extension, you may have trouble showing this; in particular, the representation is not canonical: you need to select a preimage for every $h\in H$. – Arturo Magidin May 22 '12 at 19:34
I'm sorry if your comment is unclear to me... (I'm definitely in the most general case, including split extensions but not only) The fact that the representation is non-canonical does not bother me much, but are you saying that my "proof" is wrong ? I am selecting a preimage for every $h$... – JeanThiviers May 22 '12 at 19:42
Sorry, I didn't see your answer... – JeanThiviers May 22 '12 at 19:50
I mean that you need to be a bit more careful; the coset is not enough. What you really want to do is use the $z$ coordinate to "measure" how far your $g$ is from the corresponding pre-selected $x$. See my answer; in order to really get uniqueness, you need to explain how you associate $g\in G$ with a pair. – Arturo Magidin May 22 '12 at 19:50
No problem; but as I said, you really need to explain how you are associating pairs to elements, because you need to explain how you interpret a pair $(z,h)$ as an element of $G$. When you have a split extension, there is a "natural" way of interpreting the pair as an element of $G$, namely, $(z,h)$ corresponds to $zh$. But when the extension is not split, you need to say what element of $G$ corresponds to the pair $(z,h)$, as there are many ways of making the association, and the uniqueness will be "unique relative to that association." – Arturo Magidin May 22 '12 at 19:52
up vote 1 down vote accepted

The representation is not canonical if $G$ is not a split extension.

To get a representation, let $\pi\colon G\to H$ be the canonical projection, and for each $h\in H$ select a $g_h\in \pi^{-1}(h)$.

We then associate to every $g\in G$ the element $(gg_{\pi(g)}^{-1},\pi(g))$.

Note. This is not the only way to associate elements of $G$ with pairs $Z\times H$. But in order to establish any sort of uniqueness, you need to fix an association. In fact, there are other ways of making the association (e.g., we could associate $g$ to $(g_{\pi(g)}g^{-1},\pi(g))$), so the fact that $Z$ is the kernel and induces a partition is not enough to guarantee "uniqueness". You really need to take into account how you are making pairs and elements correspond to one another.

Note that $gg_{\pi(g)}^{-1}\in Z$, since $\pi(g_{\pi(g)}) = \pi(g)$. Thus, this is an element of $Z$.

Let $(z,h)\in Z(H)$. I claim that there is one and only one element of $G$ that is associated to $(z,h)$ under the map above. First, note that $zg_{h}$ maps to the desired pair: for $\pi(zg_h) = \pi(g_h) = h$, so $(zg_h)g_{\pi(zg_h)}^{-1} = zg_hg_{h}^{-1} = z$. So we do indeed get $(z,h)$.

Now, suppose that $g$ is mapped to $(z,h)$. Then $\pi(g) = h$, and $z=gg_{\pi(g)}^{-1} = gg_{h}^{-1}$. Therefore, $g = zg_h$, as desired.

If $G$ is a split extension and $H$ is identified with a specific subgroup of $G$ (that is, if we have $G$ expressed as a semidirect product $Z\rtimes H$, then we may let $g_h = h$ for each $h\in H$ and we obtain a "natural" representation. But in general, different choices of preimages yield different representations.

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