Beckmann Rearrangement in cyclic compounds

Consider 1-indanone:

I am supposed to find the rearrangement product on treating this with hydroxylamine (which gives an oxime) and then $$\ce{H+}$$ (with heat). I am aware of the mechanism of Beckmann Rearrangement, and also know that the migration of the group anti to the hydroxyl group predominates. However, I find it difficult to predict the rearranged product as the structure is cyclic. I might be gravely wrong but If the group anti to the hydroxyl has to migrate, wouldn't the ring have to cleave? I cannot visualise any other way for this to happen. Any help would be appreciated.

• Surprise! It doesn't. There's just an additional NH group adjacent to the carbonyl in place of the carbon atom in the cyclopentane ring. (Amide) formed without cleavage. – chocolateMousse Apr 24 at 17:11
• Cyclic has nothing to do with the rearrangement. Draw each of the oximes and do the reaction as you would for an acyclic oxime. Because the rearrangement is stereospecific, the answer depends one which oxime you are rearranging. – user55119 Apr 24 at 19:20
• Without cyclic oximes, expanding to lactames, access to $\epsilon{}$-caprolactam and eventually nylon would not this significant today. – Buttonwood May 2 at 15:15