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So according to my lecture notes, the explicit form of the product of two elements in $\mathbb{Z}_{31} \rtimes \mathbb{Z}_5$ is:

$$(\mathrm{cl}_{31}(x), \mathrm{cl}_5(a)) \cdot (\mathrm{cl}_{31}(y), \mathrm{cl}_5(b)) = (\mathrm{cl}_{31}(x + 16^a y), \mathrm{cl}_5(a+b)) $$

…but I don’t really know how this is calculated. I just know that:

$$(\mathrm{cl}_{31}(x), \mathrm{cl}_5(a)) \cdot (\mathrm{cl}_{31}(y), \mathrm{cl}_5(b)) = (\mathrm{cl}_{31}(x) \phi(\mathrm{cl}_5(a))(\mathrm{cl}_{31}(y)), \mathrm{cl}_5(ab)) \,. $$

Any help would be appreciated.

Jendrik Stelzner
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Gokimo
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    The notation $cl_{n}(x)$ for the congruence class of $x$ modulo $n$ is clunky and obscuring. You'd be better off with some other notation, such as $[x]_n$, $\overline{x}$, or just $x$ with meaning understood from context. – Arturo Magidin Jan 03 '23 at 19:39

1 Answers1

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The group $\mathbb{Z}_{31}$ has an automorphism group of order $30$, given by sending $1$ to any integer between $1$ and $31$; this corresponds to the multiplicative group of invertible elements modulo $31$. This is well-known to be cyclic.

You want an automorphism of order $5$ to correspond to the action of $\mathbb{Z}_5$. Either a bit of trial and error, or knowing that $2^5=32\equiv 1\pmod{31}$ will tell you that the automorphism of $\mathbb{Z}_{31}$ given by multiplication-by-$2$ has order $5$. The other automorphisms of order $5$ are then multiplication-by-$4$, multiplication-by-$8$, and multiplication-by-$16$ (since any nontrivial element of the cyclic group of order $5$ will generate the group). It is clear that the automorphism $\phi$ that your first formula is using is the latter, multiplication-by-$16$. So $\phi(a)$ at $y$ evaluates to "multiply $y$ by $16$ and do this $a$ times", resulting in $16^ay$, taken modulo $31$.

Note that your second formula is using muliplicative notation for your groups, but $\mathbb{Z}_{31}$ and $\mathbb{Z}_5$ are groups under addition. The formula, translated into additive notation, should be (translated into your notation, with by the way is not very good...) $$(cl_{31}(x),cl_5(a)) \cdot (cl_{31}(y),cl_5(b)) = \bigl(cl_{31}(x)+\phi(cl_5(a))(cl_{31}(y)),cl_5(a+b)\bigr)$$

Note that you could have used any of the order $5$ automorphisms of $\mathbb{Z}_{31}$. Although the formula will be different (replacing $16^a$ with either $2^a$, $4^a$, or $8^a$), the resulting groups are all isomorphic to each other.

Arturo Magidin
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  • Thank you! I don’t like the notation either, but I’m trying to get used to it since it’s the one my lecturer uses. Just a question: if I understood correctly from what you wrote, $(cl_{31}(x),cl_5(a)) \cdot (cl_{31}(y),cl_5(b)) = (cl_{31}(x+2^ay),cl_5(a+b)) $ for instance would have worked as well as the formula I gave? – Gokimo Jan 03 '23 at 19:58
  • @Gokimo: Yes, it would be a group that is isomorphic to the one you gave. This is discussed, for instance, here. – Arturo Magidin Jan 03 '23 at 20:07