abstract Algebra (group theory) Three coins are placed on a table; showing heads. Can you get all the coins to show tails, by turning over two coins at a time? Use Group Theory to prove your answer. 

I know that the answer is no I cannot get all coins to show tails however I  have no idea how to put it in terms of group theory. Any help would be appreciated.
You insist on using group theory to hide the parity check, so here it is.

In group-theoretic terms the situation is as follows: we have the group $G={\bf Z}_2^3$, and the subgroup $\langle(1,1,0),(1,0,1),(0,1,1)\rangle=H\leq G$, and we ask if $(1,1,1)\in H$.

The answer is no: consider the homomorphism $f\colon G\to {\bf Z}_2$ defined by $f(a,b,c)=a+b+c$. Clearly $H$ is contained in the kernel, while $(1,1,1)$ isn't, so $(1,1,1)\notin H$.

Alternately, you could just compose the generators of $H$ in all possible ways to see that actually $H=\{(0,0,0),(1,1,0),(1,0,1),(0,1,1)\}$, though this method doesn't generalise to a larger number of coins quite so well.
Group theory has nothing to do with this, it's a simple parity check.

There are three possibilities at each step:


you turn two heads into two tails: number of heads drops by 2, number of tails goes up by 2.
you turn two tails into two heads: number of tails drops by 2, number of heads goes up by 2.
you turn a head into a tail and a tail into a head: number of heads and tails is unchanged.


Since the numbers change by leaps of 2, if you start with an odd number of tails, you still have an odd number of tails after any whatsoever sequence of changes as before.

Obviously this applies to any odd numbered set of coins you decide to start with.