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I have been working on this for several hours, and feel like I have reached the point where I am overcomplicating and confusing myself.

I know that since $\mathbb{Z}_4$ is cyclic it has automorphisms:

$$f(x)=x \mod 4$$ and $$ g(x)=3x \mod 4.$$

I'm just not sure how to define a homomorphism that will map things in $H$ to these two functions? I've tried,

$$\theta(h)=f(h)$$

and $$\theta(h) = g(h)$$

but I can't convince myself that these are homomorphisms? When we take $\theta(h_1h_2)=f(h_1h_2)$, I get stuck because $h_1h_2$ isn't in $x$, so how do we know what $f$ of it is?

In addition to this, I am also trying to define the action via automorphism for these $\theta$ and take the semidirect product of H and G, but I feel like my results when using this $\theta$ aren't groups, because they don't seem to have an identity element.

Anywho, any guidance or help would be greatly appreciated.

Greg Nisbet
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Mike
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    Any chance you reversed your $H$ and $G$? The way you currently have it, $Aut(G)$ is a group of order $2$ - it's a bit of a challenge (euphemism) to have a non-trivial homorphism from a group of order $5$ into it. – peter a g Apr 18 '21 at 03:57
  • @peterag No, the H and G are correct. I think this is what is tripping me up though. Would the map $\theta(h)=0$ be the trivial map? I wasn't sure if this would count as a homomorphism from $H$ to $Aut(G)$ or not. – Mike Apr 18 '21 at 04:02
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    I think you'll have an easier time thinking of $Aut(G)$ as being the multiplicative group ${\pm 1}$. That said, for any group $H$ there is always the trivial homomorphism $h\mapsto 1$. – peter a g Apr 18 '21 at 04:05
  • Sorry, I'm still a bit new to this group theory stuff, why can we use -1 here? – Mike Apr 18 '21 at 04:12
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    There is only one homomorphism from a group of $5$ elements to a group of $2$ elements. This is a consequence of Lagrange's theorem. Can you see why? – CyclotomicField Apr 18 '21 at 04:14
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    @Mike Sorry, I was off for the evening - but look at Chris's answer for instance. Still, to answer your last question: your $g$ is multiplication by $3$. This is the same as multiplication by $-1$ on $\mathbb Z_4$. Likewise $f$ is multiplication by $1$.This identification makes sense 'group theoretically': $g\circ g =f $ is the same as $3^2 \equiv 1 \equiv (-1)^2 \pmod 4$. So we can think of the group ${f, g}$ as the multiplicative group ${1,-1}$. – peter a g Apr 18 '21 at 14:19
  • Also, I failed to read what you tried, in your original question. If I understand, I think you're also a bit confused about $\theta$. Your title has $\theta\colon H \to Aut(G)$, while, in your text, you write $\theta(h)= f(h)$. But your $f$ is a function on $G$, so if you write $f(h)$, it looks like you want that $h$ should belong to $G$? So that is confusing.... – peter a g Apr 18 '21 at 14:27

1 Answers1

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As reported in the comments, there's only one.

Since $\varphi(4)=2$, we have $\rm{Aut}(G)\cong\mathbb Z_2$. That's because of the general fact that $\rm {Aut}(\mathbb Z_n)\cong \mathbb Z_n^×$.

So we are looking for homomorphisms from $\mathbb Z_5$ to $\mathbb Z_2$.

But $(5,2)=1$, and by Lagrange and the homomorphism property, $|h(1)|\mid5$ and $|h(1)|\mid2$.
$\therefore |h(1)|\mid1 $.

So there is only the trivial one.