# 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

Staff Emeritus
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.