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I am trying to write down the central series of $D_4$. i.e $e$=$Z_0$ $\subset Z_1 \subset Z_2........\subset Z_l=D_4$ , where $Z_1=Z(D_4)$ and $Z_{i+1} $ is defined such that $Z_{i+1}/Z_i=Z(G/Z_i)$, where $Z$ denotes the center.

This is what i did, we know that the centre of $Z$ is ${e, r^2}$. to find $Z_2$, the i found the centre of the factor group $G/Z_i $ which is the modulo class of two elements $(e,r^2), (r,r^3)$ . I am not sure if i am right or my procedure is right . I need some tips . Thanks

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    @Lubin : Is it exactly how central series is determined ? I am not sure if i am understanding . that means i have $e \subset {(e,r^2)} \subset D_4$2012-12-10

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That's exactly what you do. You know $Z_1=Z(D_4)$, so to find $Z_2$, first find $Z_2/Z_1=Z(D_4/Z_1)$. Then you take the "lift" of $Z_2/Z_1$ by the correspondence theorem - that is, you figure out which of the subgroups of $D_4$ that contain $Z_1$ correspond to the subgroup $Z_2/Z_1$ of $D_4/Z_1$. (See: correspondence theorem.) In the general case doing this can occasionally be tricky, but most of the time it's achievable with a simple order argument. In this case, the argument is simply that $|Z_2/Z_1|=4$ and $|Z_1|=2$ so $|Z_2|=8$, whence $Z_2$ is all of $D_4$.