Recent content by astronut24

Hello need help with this one. f:[0,1] --> [0,1] f( .x1 x2 x3 x4 x5 ...) = .x1 x3 x5 x7 ( decimal expansion) prove that f is nowhere diffrentiable but continuous. i tried by just picking a point a in [0,1] and the basic definiton of differentiability about that point...doesnt seem...

if G is a group such that a^2=e for each a in G. show that order of G is 2^n for some n>=0. please help... is the group of 3 non-singular upper triangular matrices a normal subgroup of GL(3,R), the group of 3 cross 3 non singular matrices over R.

thanks for the help! here's another question on the same lines: if G is a group where a^2=e for each a in G. show that order of G is 2^n for some n >=0. it's clear that the group is abelian and also clear that 2 divides O(G). how do you proceed further? by any chance, is it induction that we're...

if a group of order 2p ( p prime) is abelian...then does it have exactly one element of order 2 ?? if a group is non abelian...i could figure out that there are p elements of order 2. but the abelian case is a bit confusing... also..is it like...any group of order 2p has an element of order p...

if G is a group of order 2n then show that it has an element of order 2 ( and odd number of them) i've been thinking about this...and i think I've gotten somewhere... e belongs to G and o(e) = 1 now if there's no element of order 2 in G...we're looking at elements which are not their own...

another question... well...thanks...the order of xax^-1 is 2...and a is the only element with order 2...so xax^-1=a and that implies the result. another problem that i seem to be unable to figure out is... f:(Zm , +m) --> (Zn,+n) is a group homomorphism where Zm and Zn denote groups of...

i've just started out with a course in group theory...here's a question that's been bothering me for a while now... let G be a group and 'a' ,a unique element of order 2 in G. show that a belongs to Z(G). if every element of the group has order 2 this is pretty easy...but that's not the case...

ya...so what further? that's a valid point you've raised... so do you divide by 4?

hello well...yes, it's possible to have a cycle of length 2 in addition to the 5 cycle...but then the l.c.m becomes 10. so that rules out such a consideration!

what is the number of elements of order 5 in the permutaion group S7?? so what we're concerned with here is, after decompositon into disjoint cycles the l.c.m of the lengths must be 5. since 5 is a prime, the only possible way we could get 5 as l.c.m would be to fix ANY 2 elements amongst the 7...