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Find isomorphic refinements of the series: $ \left \{ 0\right \} \triangleleft \langle 12 \rangle \triangleleft \langle 3 \rangle \triangleleft \mathbb{Z}_{48}$

My idea of what a refinement actually is, is a little vague and I feel that if I were to see an example of how one is found my understanding will become clearer. Any help is much appreciated.

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$$\langle 12\rangle=\{0,12,24,36\}\;\;,\;\;\langle 3\rangle =\{0,3,6,9,12,15,18,21,24,27,30,33,36,39,42,45\}\Longrightarrow$$

$$\left|\langle 12\rangle/\{0\}\right|=\left|\langle 12\rangle\right|=4\,\,,\,\,\left|\langle 3\rangle/\langle 12\rangle\right|=\frac{16}{4}=4\,\,\,,\,\,\left|\Bbb Z_{48}/\langle 3\rangle\right| = 3$$

and, of course, everything's cyclic above , so for example of refinement we can take

$$\{0\}\triangleleft \langle 12\rangle\triangleleft\langle 6\rangle\triangleleft\langle 3\rangle\triangleleft\Bbb Z_{48}$$

The calculation of the orders above was to make things easier.

You try now to find another refinement with isomorphic factors as the last one...can you see where "to stick" a subgroup in between the original series?

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