Homework Help: Basic Group Theory Question

1. Jun 6, 2014

PsychonautQQ

1. The problem statement, all variables and given/known data
Let (R^2,+) be the set of ordered pairs with addition defined component wise. Verify {(x,2x)|x£R} is a subgroup and that {(x,2x+1)|x£R} is not a subgroup.

3. The attempt at a solution
So for something to be a subgroup it has to have all it's set items contained in the main group and be a group on it's own.

To be a group on it's own it has to have an identity e and an operation (in this case +) so that:
e+g = G

and an inverse so that (in this case a negative number with same magnitude)
g + (-g) = 0

and be assosciative
(g+h)+k = g+(h+k)
and it seems both (x,2x) and (x,2x+1) satisfy all these properties.

Furthermore, it seems that both (x,2x) and (x,2x+1) have all their combinations included in the main group. How is (x,2x+1) not a subgroup? What am I missing here?

2. Jun 6, 2014

HallsofIvy

I presume you mean e+ g= g.

Since you are using ordinary addition, (x, 2x+ 1)+ (y, 2y+ 1)= (x+ y, 2x+1+ 2y+ 1)= (x+ y, 2(x+ y)+ 2) which is NOT of the form (a, 2a+ 1).

3. Jun 6, 2014

AlephZero

You are missing the fact that if H is a subgroup of G, and x and y are any elements of H, then x+y must also be an element of H.

4. Jun 6, 2014

pasmith

The operation of the subgroup is the same as the operation of the full group. Thus you already know that the operation is associative, that there is a unique identity $e \in G$, and that every $g \in G$ has a unique inverse $g^{-1} \in G$.

To show that $H \subset G$ is a subgroup, you must show the following (in order of being easiest to check):
• Identity: $e \in H$.
• Inverse: If $h \in H$ then $h^{-1} \in H$.
• Closure: If $h_1 \in H$ and $h_2 \in H$ then $h_1h_2 \in H$.

Here $H = \{(x, 2x + 1) : x \in \mathbb{R}\}$ falls at the first hurdle: $(0,0) \notin H$.

5. Jun 6, 2014

Zondrina

Pasmith has noted something important in his post. If the origin isn't included, then it can't be a subgroup. Any line with a y intercept that isn't 0 can't form a subgroup.

So $(x,2x)$ will form a subgroup, but $(x,2x + b)$ for any $b ≠ 0$ will not.