# Is the Mod2-Reduction Map Onto?

1. Aug 2, 2011

### Bacle

Hi, Algebraists:

The modN reduction map r(N) from a matrix group (any group in which the elements
are matrices over Z-integers) over the integers, in which r is defined by

r(N) : (a_ij)-->(a_ij mod N) is not always commutative, e.g.:

r(6) :Gl(2,Z) --Gl(2,Z/N)

is not onto, since Gl(2,Z) is unimodular over Z, but Gl(2,Z/N) is not, e.g., we can

take units of Z/N that are not 1modN , so that the determinants do not match up;

e.g., for N=10 , take the unit , say, 7 in Z/10Z ; then the matrix M with a_11=7

a_22 =0 and 0 otherwise, is in Gl(2,Z/10) (use, M' with a_11'=3 , and a_22'=1 )

but it is not the image of any matrix in Gl(2,Z), since the determinants do not match

up.

** question **:

Anyone know any results re when the mod2 reduction map r(2) is onto?

Thanks.

2. Dec 28, 2011

### Li(n)

Ughh I'm coming across a similar problem..

3. Dec 30, 2011

### Bacle2

Sorry; I still don't have a good answer. I will keep looking into it.