Extending automorphisms on cyclic subgroups

Question

Suppose G is a finite, cyclic group and H a subgroup (also cyclic by nature). Let $f:H\to H$ be an automorphism. Does there exist an automorphism $g:G\to G$ such that $g|_H = h$? That is, when restricted to H, g equals h.

Thank you.

Answer 1

Such an extension is always possible.

Suppose $G$ is a finite cyclic group, $H$ a subgroup of $G$, and suppose $h\,\colon H \rightarrow H$ is an automorphism of $H$.

The goal is to show that there is an automorphism $g\,\colon G \rightarrow G$ which extends $h$.

If $H$ is the trivial group, let $g$ be the identity map on $G$.

If $H = G$, let $g = h$.

Assume then that $H$ is a nontrivial proper subgroup of $G$.

Let $n = o(G)$ and let $m = o(H)$. Then $m|n$ and $1 < m < n$.

Without loss of generality, represent $G$ by $Z_n = \{0,1,...,n-1\}$.

Since $G$ is cyclic, so is $H$.

Let $k = n/m$.

As a subgroup of $G$, we must have $H = \{0,k,...,k(m-1)\}$.

Since $h$ is an automorphism of $H$, it follows that $h$ can be expressed in the form $h(x) = ax$, for some positive integer $a < m$ with $\text{gcd}(a,m)=1$.

Let $W = \{0,1,...,k-1\}$. For $w \in W$, let $b_w = a + wm$ and let $g_w\,\colon G \rightarrow G$ be defined by $g_w(x) =b_wx$.

Then for all $w \in W$, $g_w$ extends $f$.

Furthermore, for a given $w \in W$, $g_w$ is an automorphism of $G$ if and only if $\text{gcd}(b_w,n) = 1$.

Our goal is to show $\text{gcd}(b_w,n) = 1$ for some $w \in W$.

Since $\text{gcd}(a,m) = 1$ , it follows that $\text{gcd}(b_w,m) = 1$ for all $w \in W$.

Let $j$ be the largest positive integer factor of $n$ such that $\text{gcd}(j,m) = 1$.

Suppose first that $j = 1$. Then all prime factors of n are also prime factors of m, hence, since $\text{gcd}(b_0,m) = 1$, it follows that $\text{gcd}(b_0,n) = 1$, and we're done.

Next suppose $j>1$.

Since $\text{gcd}(j,m) = 1$ and $n = mk$, necessarily $j|k$.

Let $V = \{0,...,j-1\}$. Then $j|k$ implies $j \le k$, hence $V \subseteq W$.

Claim no two of $b_0,...,b_{j-1}$ are congruent, mod $j$.

Suppose $b_r \equiv b_s \pmod j$, where $0 \le r < s \le j-1$. Then

\begin{align*} &b_r \equiv b_s{\pmod j}\\[6pt] \implies\; &a + rm \equiv a + sm{\pmod j}\\[6pt] \implies\; &j|(s-r)m\\[6pt] \implies\; &j|(s-r)\\ \end{align*}

contradiction, since $0 < s-r < j$.

Since $b_0,...,b_{j-1}$ are distinct, mod $j$, it follows $b_v \equiv 1 \pmod j$, for some $v \in V$.

Then $\text{gcd}(b_v,j) = 1$ and $\text{gcd}(b_v,m) = 1$ implies $\text{gcd}(b_v,n) = 1$, hence $g_v$ is an automorphism of $G$ which extends $h$.

This completes the proof.