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math.stackexchange.com/questions/1302721/is-t … mautg-is-abelian-but
If $G$ is finite abelian and not cyclic, it can be written as direct sum $G=G_1\oplus G_2\oplus\ldots \oplus G_n$ of $n\ge2$ nontrivial cyclic groups of orders dividing each other. Say $G_1\cong \mathbb Z/m\mathbb Z$ and $G_2\cong \mathbb Z/md\mathbb Z$. As $G_1\oplus G_2$ is a direct summand, $\operatorname{Aut}(G_1\oplus G_2)$ is a subgroup of $\operatorname{Aut}(G)$.
The group $\mathbb Z/m\mathbb Z\oplus\mathbb Z/md\mathbb Z$ has especially the automorphisms $$\phi\colon (x+m\mathbb Z,y+md\mathbb Z)\mapsto (x+y+m\mathbb Z,y+md\mathbb Z)$$ and $$\psi\colon(x+m\mathbb Z,y+md\mathbb Z)\mapsto(x+\mathbb Z,y+dx+md\mathbb Z).$$
We have
$\psi(\phi(m\mathbb Z,1+md\mathbb Z))=(1+m\mathbb Z,1+d+md\mathbb Z)$ and $\phi(\psi(m\mathbb Z,1+md\mathbb Z))=(1+m\mathbb Z,1+md\mathbb Z)$. As $m>1$, this shows that $\operatorname{Aut}(G_1\oplus G_2)$ is not abelian.
$$(mℤ,1+mdℤ)\oversetϕ→(1+mℤ,1+mdℤ)\oversetψ→(1+mℤ,1+d+mdℤ)$$$$(mℤ,1+mdℤ)\oversetψ→(mℤ,1+mdℤ)\oversetϕ→(1+mℤ,1+mdℤ)$$当$m>1$时不同构.
去掉“有限群”的限制,结论不成立,见this answer
In general however, a group with abelian automorphism group need not be abelian, i.e., in particular not cyclic. See the article
Jonah, D.; Konvisser, M.
Some non-abelian $p$-groups with abelian automorphism groups.
Arch. Math. (Basel) 26 (1975), 131--133.
In the infinite case, there are locally cyclic groups that are not cyclic, and these have abelian automorphism groups. For instance, the additive group of rational numbers has an abelian automorphism group (the multiplicative group of rational numbers). |
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