Why is the Automorphism group of cyclic group of order of power eof 2 not just $Z 2^{n-1}$

Question

I'm banging my head trying to figure out why the automorphism group of group of order $2^n$ not just the cyclic group of $2^{n-1}$, isn't the automorphism completely determined by where $e$ is sent to? in this case only the odd number are possible so it should just be $2^{n-1}$ right?

Answer 1

The automorphism group has order $2^{n-1}$ certainly. But it's not cyclic in general. If $n\ge 3$ there are automorphisms taking a generator $g$ to $g^{-1}$ and also taking $g$ to $g^{1+2^{n-1}}$. Each of these automorphisms has order $2$.

Answer 2

The given group can and will be taken (up to isomorphism) as $(\Bbb Z/2^n,+)$, the underlying group of the ring $$ R = (\Bbb Z/2^n,+,\cdot)\ ,$$ and we can identify the automorphism $\phi_a:(R,+)\to(R,+)$, determined by $1\to a$, $a$ odd representative among $1,3,\dots, 2^n-1$ modulo $2^n$, with the element $a\in R^\times$. The composition $\phi_a\circ\phi_b=\phi_b\circ\phi_a$, $a,b\in R^\times$, sends $$1 \to a=\underbrace{1+\dots+1}_{a\text{ times}} \to \underbrace{b+\dots+b}_{a\text{ times}} =ab $$ (modulo $2^n$), so we can identify the automorphism group of $\Bbb Z/2^n$ with $(R^\times ,\cdot)$. It turns out (see the final part marked as "comupter support" for this) that for all $n\ge 3$ the generators are

• $-1$ of order $2$
• and $5$ of order $2^{n-2}$. (This is the only new information, compared with the answer of Lord Shark the Unknown.)

We consider from now on only the case $n\ge 3$, then:

Claim: $$ \begin{aligned} \operatorname{Aut}(\ (\Bbb Z/2^n,+)\ ) &\cong (\ (\Bbb Z/2^n)^\times\ ,\ \cdot\ ) \\ &=\langle\ -1,5 \text{ in } (\Bbb Z/2^n)^\times\ \rangle \\ &\cong (\Bbb Z/2,+)\times(\Bbb Z/2^{n-2},+)\ . \end{aligned} $$

Proof: It is clear that $-1$ has order two.

Also, the equation $x^2=1$ in $R$ has also the solutions $2^{n-1}\pm 1=\pm(2^{n-1}+1)$, easily checked by squaring.

For the structural part it is enough to show that there are only these three elements of order two in $R^\times$, which together with $1$ are forming the Klein group $K=(\ \{\pm 1,\pm 2^{n-1}\} \cdot\ )$. This is easy, let $x=2^ks+1$ with an odd $s$, and $k\ge 1$, be an element of order two. For $n\ge 3$, in case of $k\ge 2$ we replace $x$ by $-x$, to get an element of order two with $k=1$, $x=2s$.

Then $x^2=1$

• is equivalent to $(x-1)(x+1)=0$,
• i.e. $2s(2s+2)=0$,
• i.e. $4(s+1)=0$,
• i.e. $s+1\in \{0,\pm 2^{n-2},2\cdot 2^{n-2}\}$,
• i.e. $x=2s+1\in \{-1,2^{n-1}-1\}$.

The 2-torsion is now cleared, the Klein group $K$ above.

Now we observe that there is no square root of $-1$ in $R$, i.e. working modulo $2^n$, since there is none modulo $4$.

Since $R^\times$ has the structure $\cong \Bbb Z/2^s\times \Bbb Z/2^t$, with $2^s2^t=|R^\times|=2^{n-1}$, we already have $s=1$, $t=n-1$ (or conversely), and the structural, existential information of an isomorphism as in the Claim. Now we put the hands explicitly on the generator of the part $\Bbb Z/2^{n-1}$. Either $3$, or $5$ is a generator. Indeed, working in $R$, $n\ge4$ (to have a known sign below, the case $n=3$ is checked manually):

$$ \begin{aligned} (4\pm1)^{2^{n-1}} &=(4\pm1)^{\phi(2^n)}=1\ , \\ &\qquad\text{ using the Euler indicator, but even better} \\ (4\pm 1)^{2^{n-2}} &= 1\ , \\ &\qquad\text{ working in $R^\times/\pm1$, but even better} \\ (4\pm 1)^{2^{n-3}} &= (\pm1) -4\begin{pmatrix}2^{n-3}\\ 1\end{pmatrix} \underbrace{ +4^2\begin{pmatrix}2^{n-3}\\2\end{pmatrix} -4^3\begin{pmatrix}2^{n-3}\\3\end{pmatrix} +4^4\begin{pmatrix}2^{n-3}\\4\end{pmatrix} \pm \dots}_{=0} \\ &=\pm 1 +2^{n-1}\ , \end{aligned} $$ and this determines the needed order.

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Computer support:

Here is the structure of this group of units for the first values of $n$:

sage: for n in [1..6]:
....:     R = Zmod(2^n)
....:     A = R.unit_group()
....:     generators = [ R(gen) for gen in A.gens() ]
....:     print ( "n = %s\n\t(R+,.) is %s\n\t(R*,.) is %s\n\tGenerators %s"
....:             % ( n, R, A, generators ) )
....:     
n = 1
        (R+,.) is Ring of integers modulo 2
        (R*,.) is Trivial Abelian group
        Generators []
n = 2
        (R+,.) is Ring of integers modulo 4
        (R*,.) is Multiplicative Abelian group isomorphic to C2
        Generators [3]
n = 3
        (R+,.) is Ring of integers modulo 8
        (R*,.) is Multiplicative Abelian group isomorphic to C2 x C2
        Generators [7, 5]
n = 4
        (R+,.) is Ring of integers modulo 16
        (R*,.) is Multiplicative Abelian group isomorphic to C2 x C4
        Generators [15, 5]
n = 5
        (R+,.) is Ring of integers modulo 32
        (R*,.) is Multiplicative Abelian group isomorphic to C2 x C8
        Generators [31, 5]
n = 6
        (R+,.) is Ring of integers modulo 64
        (R*,.) is Multiplicative Abelian group isomorphic to C2 x C16
        Generators [63, 5]

sage continues to give the same generators, $-1$ and $5$ for the "next few" values of $n$. For instance:

n = 99
        (R+,.) is Ring of integers modulo 633825300114114700748351602688
        (R*,.) is Multiplicative Abelian group isomorphic to C2 x C158456325028528675187087900672
        Generators [633825300114114700748351602687, 5]
n = 100
        (R+,.) is Ring of integers modulo 1267650600228229401496703205376
        (R*,.) is Multiplicative Abelian group isomorphic to C2 x C316912650057057350374175801344
        Generators [1267650600228229401496703205375, 5]