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I want to show for $n\in\Bbb N$, which is not square-free, that every subgroup of $Z_n=\langle x\;|\;x^n\rangle$ is characteristic. But I want to show it in a convoluted way. Every automorphism $\varphi$ of a group $G$ induces a lattice automorphism $\varphi^L$ on the subgroup lattice $L(G)$. This lattice automorphism is defined by $\varphi^L=(H\mapsto\varphi(H))$. We indeed have that $\varphi^L(H\cap K)=\varphi^L(H)\cap\varphi^L(K)$ and $\varphi^L(H\vee K)=\varphi^L(H)\vee\varphi^L(K)$. This can be extended to show that $\varphi^L$ is in fact a complete lattice automorphism, but that's not pertinent to this question since every finite lattice is complete.

If I can show that $L(Z_n)$ has only one automorphism (the identity), this will show that every subgroup of $Z_n$ is characteristic since every automorphism of $Z_n$ would induce the trivial automorphism on $L(Z_n)$.

Perhaps now, you can see why I excluded the square-free case, since for square-free $n$ (with at least 2 prime divisors) the subgroup lattice of $Z_n$ is isomorphic to the power-set lattice of some set. Namely, if $n=p_1 p_2\cdots p_k$ then $L(Z_n)$ is isomorphic to the power-set lattice of a $k$-element set. If $n$ is square-free, then a permutation of the atoms of $L(Z_n)$ would induce a non-trivial lattice automorphism.

So, the main question is:

For $n\in\Bbb N$ which is not square-free, is it true that $L(Z_n)$ has only one lattice automorphism (the identity)?

If $n=p^k$, then $L(Z_n)$ is a chain of length $k+1$, and every finite chain has only one lattice automorphism.

If $n=p_1^{e_1}\cdots p_k^{e_k}$, I know that the atoms of $L(Z_n)$ are of the form $\langle x^k\rangle$ where $k=n/p_i$ for some $p_i$ in the prime-factorization of $n$. However, $L(Z_n)$ need not be atomistic just by considering $Z_{12}$, so it would not suffice to show that every atom is fixed by a lattice automorphism.

So, I'm stumped. Any help is appreciated.

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  • $\begingroup$ "But I want to show it in a convoluted way." Why?!?! $\endgroup$ – Pedro Tamaroff Aug 31 '14 at 2:55
  • $\begingroup$ A part of math is about discovering new ways to prove already known facts. Unfortunately in this case, James has provided a counter-example. $\endgroup$ – Robert Wolfe Aug 31 '14 at 3:00
  • $\begingroup$ @Bryan. Absolutely! It is always worthwhile trying to draw connections between various, perhaps apparently disparate ideas. Even if an idea doesn't work out the way you had hoped, the process of working it through will teach you something. $\endgroup$ – James Aug 31 '14 at 6:28
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It seems that the subgroup lattice of the cyclic group $C_{36}$ has non-trivial lattice automorphisms. From the lattice diagram

Subgroup lattice of cyclic group of order 36

of the subgroups of $C_{36}$, you can see that the mapping $$7\leftrightarrow 8, \; 4\leftrightarrow 6, \; 2\leftrightarrow 3,$$ leaving the other subgroups fixed, is a non-trivial lattice automorphism. Of course, the automorphism cannot have been induced by an automorphism of the underlying group, at least in this case.

The subgroups (labelled with numbers to easily describe the lattice automorphism) in the lattice diagram have the following orders:

  1. $1$
  2. $2$
  3. $3$
  4. $4$
  5. $6$
  6. $9$
  7. $12$
  8. $18$
  9. $36$
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