Editing Alternating group (section)

==Group homology==
{{see also|Symmetric group#Homology}}
The [[group homology]] of the alternating groups exhibits stabilization, as in [[stable homotopy theory]]: for sufficiently large ''n'', it is constant. However, there are some low-dimensional exceptional homology. Note that the [[Symmetric group#Homology|homology of the symmetric group]] exhibits similar stabilization, but without the low-dimensional exceptions (additional homology elements).

=== ''H''<sub>1</sub>: Abelianization ===
The first [[homology group]] coincides with [[abelianization]], and (since A<sub>''n''</sub> is [[perfect group|perfect]], except for the cited exceptions) is thus:
:''H''<sub>1</sub>(A<sub>''n''</sub>, Z) = Z<sub>1</sub> for ''n'' = 0, 1, 2;
:''H''<sub>1</sub>(A<sub>3</sub>, Z) = A{{su|b=3|p=ab|lh=1em}} = A<sub>3</sub> = Z<sub>3</sub>;
:''H''<sub>1</sub>(A<sub>4</sub>, Z) = A{{su|b=4|p=ab|lh=1em}} = Z<sub>3</sub>;
:''H''<sub>1</sub>(A<sub>''n''</sub>, Z) = Z<sub>1</sub> for ''n'' ≥ 5.

This is easily seen directly, as follows. A<sub>''n''</sub> is generated by 3-cycles – so the only non-trivial abelianization maps are {{nowrap|A<sub>''n''</sub> → Z<sub>3</sub>,}} since order-3 elements must map to order-3 elements – and for {{nowrap|''n'' ≥ 5}} all 3-cycles are conjugate, so they must map to the same element in the abelianization, since conjugation is trivial in abelian groups. Thus a 3-cycle like (123) must map to the same element as its inverse (321), but thus must map to the identity, as it must then have order dividing 2 and 3, so the abelianization is trivial.

For {{nowrap|''n'' < 3}}, A<sub>''n''</sub> is trivial, and thus has trivial abelianization. For A<sub>3</sub> and A<sub>4</sub> one can compute the abelianization directly, noting that the 3-cycles form two conjugacy classes (rather than all being conjugate) and there are non-trivial maps {{nowrap|A<sub>3</sub> ↠ Z<sub>3</sub>}} (in fact an isomorphism) and {{nowrap|A<sub>4</sub> ↠ Z<sub>3</sub>}}.

=== ''H''<sub>2</sub>: Schur multipliers ===
{{main|Covering groups of the alternating and symmetric groups}}
The [[Schur multiplier]]s of the alternating groups A<sub>''n''</sub> (in the case where ''n'' is at least 5) are the cyclic groups of order 2, except in the case where ''n'' is either 6 or 7, in which case there is also a triple cover. In these cases, then, the Schur multiplier is (the cyclic group) of order 6.<ref name="raw">{{citation|first=Robert |last=Wilson |author-link=Robert Arnott Wilson |date=October 31, 2006 |url=http://www.maths.qmul.ac.uk/~raw/fsgs.html |title=The finite simple groups, 2006 versions |chapter=Chapter 2: Alternating groups |chapter-url=http://www.maths.qmul.ac.uk/~raw/fsgs_files/alt.ps |postscript=, 2.7: Covering groups |url-status=dead |archive-url=https://web.archive.org/web/20110522121819/http://www.maths.qmul.ac.uk/~raw/fsgs.html |archive-date=May 22, 2011 }}</ref> These were first computed in {{Harv|Schur|1911}}.

:''H''<sub>2</sub>(A<sub>''n''</sub>, Z) = Z<sub>1</sub> for ''n'' = 1, 2, 3;
:''H''<sub>2</sub>(A<sub>''n''</sub>, Z) = Z<sub>2</sub> for ''n'' = 4, 5;
:''H''<sub>2</sub>(A<sub>''n''</sub>, Z) = Z<sub>6</sub> for ''n'' = 6, 7;
:''H''<sub>2</sub>(A<sub>''n''</sub>, Z) = Z<sub>2</sub> for ''n'' ≥ 8.