# Presentation of the holomorph of $\mathbb Z/5 \mathbb Z$

When I look up the presentation of the holomorph of $\mathbb Z/5 \mathbb Z$ it reads like the following: $\left\langle a,b \mid a^5 = 1, b^4 = 1, bab^{-1} = a^2\right\rangle$

The automorphisms of $N := \mathbb Z/5 \mathbb Z$ are as follows $Aut(N) = \{\iota, \psi, \psi^2, \psi^3\}$ where $\psi: \mathbb Z/5 \mathbb Z \to \mathbb Z/5 \mathbb Z$ via $x \mapsto 3\cdot x$ and $\iota: \mathbb Z/5 \mathbb Z \to \mathbb Z/5 \mathbb Z$ via $x \mapsto x$. This means in detail:

• $\psi(x) = 3\cdot x$
• $\psi^2(x) = -x$
• $\psi^3(x) = 2\cdot x$
• $\psi^4(x) = x$

Define the holomorph of $N$ as $G := N \rtimes Aut(N)$ via $(g_1, \eta_1)\star (g_2, \eta_2) := (g_1 + \eta_1(g_2), \eta_1\circ\eta_2)$. Note $(0,\iota)$ is the identity element in $G$.

Define (possible) generators $x := (1,\iota)$ and $y := (0, \psi)$ of $G$.

• $x^2 = (1,\iota)\star (1,\iota) = (1 + \iota(1),\iota\circ\iota) = (1 + 1,\iota\circ\iota) = (2,\iota)$
• $x^3 = (1+2,\iota) = (3,\iota)$
• $x^4 = (4,\iota)$
• $x^5 = (0,\iota)$
• $y^2 = (0, \psi) \star (0, \psi) = (0 + \psi(0), \psi\circ\psi) = (0+0,\psi^2) = (0,\psi^2)$
• $y^3 = (0, \psi^3)$
• $y^4 = (0, \psi^4) = (0,\iota)$
• $y\star x \star y^{-1} = (0, \psi)\star (1,\iota) \star (0, \psi^3) = (0+\psi(1),\psi) \star (0, \psi^3) = (3,\psi) \star (0, \psi^3) = (3+\psi(0),\psi^4) = (3,\iota)= x^3$

So, everything works out except for $y\star x \star y^{-1} =x^3$ instead of $y\star x \star y^{-1} = x^2$ as in the presentation given above.

Questions:

1. Did I make an error? (I apologize if it is obvious to you.)
2. Are the presentations $\left\langle a,b \mid a^5 = 1, b^4 = 1, bab^{-1} = a^2\right\rangle$ and $\left\langle a,b \mid a^5 = 1, b^4 = 1, bab^{-1} = a^3\right\rangle$ equivalent? If so, is there an easy argument?

• You know that the order of coposotion of functions can be written in two ways. That is, if $f$ and $g$ are functions, is $f*g$ applying $f$ before $g$ or the reverse order? – Somos May 29 '18 at 17:53
• In $f*g$ I would apply $g$ first and then $f$. Sorry, I did not think of that: it is a matter of habit - and this is of course different for everyone. In detail: $(f*g)(x) = f(g(x))$. – Moritz May 29 '18 at 17:55
• But what if you tried the reverse order? Or else, in generator $y$ use $\psi^4$ instead of $\psi$ – Somos May 29 '18 at 17:59
• Yes, to go from one to the other, replace $b$ by $b^{-1}$ (i.e by $b^3$). – Tobias Kildetoft May 29 '18 at 17:59
• @Somos: I will check it. – Moritz May 29 '18 at 18:08

Let let $x := (1,\iota)$ and $y := (0, \psi^3)$ as motivated by the comment of Tobias Kildetoft above.
• $y^2 = (0, \psi^3) \star (0, \psi^3) = (0 + \psi^3(0), \psi^3\circ\psi^3) = (0+0,\psi^2) = (0,\psi^2)$
• $y^3 = (0, \psi^5) = (0, \psi)$
• $y^4 = (0, \psi^4) = (0,\iota)$
• $y\star x \star y^{-1} = (0, \psi^3)\star (1,\iota) \star (0, \psi) = (0+\psi^3(1),\psi^3) \star (0, \psi) = (2,\psi^3) \star (0, \psi) = (2+\psi^3(0),\psi^4) = (2,\iota)= x^2$