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March 17

Zappa–Szép product

Let G be a group. Is $Sym(G)$ an internal Zappa–Szép product of G and $Sym(G\backslash \{e\})$ , where e is the identity element of G? GeoffreyT2000 (talk) 01:30, 17 March 2016 (UTC)[reply]

Technically no since neither G nor Sym(G\e) is a subgroup of Sym(G). But if you map G→Sym(G) by making g act on G using right (or left) multiplication, and identify Sym(G\e) with Sym(G)_e, then Sym(G) is an internal Z-S product of groups isomorphic to G and Sym(G\e). (That's assuming I understand the definitions in the article; it's not a concept I'm familiar with.) You could equally well say that Sym(G) is isomorphic to an external Z-S product of G and Sym(G\e). --RDBury (talk) 18:45, 17 March 2016 (UTC)[reply]

Trapezoid with perpendicular diagonals

Does there exist a trapezoid (other than a rhombus if parallelograms are considered trapezoids) with perpendicular diagonals? 63.251.215.25 (talk) 17:06, 17 March 2016 (UTC)[reply]

Yes, there are lots of them. They are easy to draw if you start with the diagonals. Here is one.Dbfirs 17:51, 17 March 2016 (UTC)[reply]

Though it doesn't directly answer the question, the article on orthodiagonal quadrilaterals may be of interest here. --RDBury (talk) 19:41, 17 March 2016 (UTC)[reply]

Note that in North American English the words "trapezoid" and "trapezium" are interchanged in meaning compared to their original¹ meanings, which are still the ones used outside North America. So the meaning of the question depends on who's asking it. But it doesn't matter here, as the answer is yes either way. Just construct a triangle ABC where A is a right angle and extend BA to D and CA to E. In North America, choose E so that DE and BC are parallel, but BE and CD are not; elsewhere, choose E so that none of the lines are parallel. Then BCDE is the trapezoid you want.

¹Think of a circus trapeze with its two parallel ropes, corresponding to the original sense of "trapezium". Then the -oid ending indicates that a trapezoid is something like a trapezium but not exactly, so it doesn't have the parallel sides. There was an influential textbook in Britain that got the words backwards, and this happened to happen at the same time that American academia was separating itself from British, so the later restoration of the original sense in Britain didn't get copied in North America. Sorry, no cite; I forget where I read this. The OED identifies the textbook as Hutton's Mathematical Dictionary (1795) but in the OED's 1st edition, which is all I have time to check now, doesn't address the usage in North America.

--69.159.61.172 (talk) 00:42, 18 March 2016 (UTC)[reply]

Yes, Charles Hutton probably got the reversed meaning from "A Commentary on the first book of Euclid’s Elements" poorly translated from Proclus in 1788 by Taylor: "Of non-parallelograms, some have only two parallel sides,..others have none of their sides parallel. And those are called Trapeziums, but these Trapezoids.". In the UK, the word trapezoid is unused except in the same sense as American usage, or for a 3-D solid with faces that have two parallel sides. It is not used for a general quadrilateral. Dbfirs 21:44, 18 March 2016 (UTC)[reply]

Going through American mathematics, the quadrilaterals whose only special characteristic was perpendicular diagonals was not even covered as a special quadrilateral. However if one of those two perpendiculars was split equally by the crossing, it was often called a "kite".Naraht (talk) 12:39, 18 March 2016 (UTC)[reply]

We do have an article on Kite_(geometry). SemanticMantis (talk) 16:31, 18 March 2016 (UTC)[reply]