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Shahla Ahdout and Sheldon Rothman

In a standard undergraduate course in linear algebra, one learns that the only subspaces of the plane are the zero subspace, the entire plane itself, and the (straight) lines through the origin. This is easily implied from the definition of a subspace which requires it to be closed under linear combinations and scalar products. However we could not have reached this conclusion by just using the fact that a subspace is also a subgroup of the plane. Consider the following example.

Example 1. Let L = \{(x,y): x \text{and} y \text{are integers} \}. Then L is clearly a subgroup of the plane that is not a line through the origin or the entire plane or the origin. Of course L is not a subspace since, for example, it is not closed under scalar multiplication.

Noticing that the lattice L described in Example 1 is not a connected subset of the plane, suppose that we impose the extra restriction of connectedness on our subgroup. Can we conclude that the subgroup is one of the special types mentioned above? Although the situation here is not as obvious, the non-trivial Example 2 at the end of the paper will show that the answer is still ``No''. An additional hypothesis is needed. In this paper we will show the following result.

A subgroup must be either a straight line through the origin or the entire plane if either

  • (1) {the subgroup is connected and contains a straight line segment, called a linear d-arc,}


  • (2) {the subgroup contains a subset of the form $\{(x,f(x))\}$ or $\{(f(x),x)\}$ where $f$ is a continuous, non-linear function from an interval into $R$. We call such a subset a non-linear d-arc. In this case, the subgroup is the entire plane.}

    Next, we ask, if in (1) above, we replace connected with the stronger condition of arcwise connected, can we eliminate the additional hypothesis of containing a linear d-arc and still draw the same conclusion? Here the answer is ``yes'' (now allowing the trivial subgroup, i.e. the origin, as a possibility) and this is a consequence of the much broader result [6]:

    Yamabe's Theorem

    An arcwise connected subgroup of a Lie group is a Lie group.

    The reader unfamiliar with the concept of a Lie group need not be concerned since the only Lie group of interest in this paper is the plane, denoted by R^2. Applied to this group, the theorem affirms one's intuition about (additive) subgroups of the plane, namely:

    Yamabe's Theorem in the Plane

    An arcwise connected (additive) subgroup of the plane must be one of the following: the trivial subgroup consisting of the origin alone, a straight line through the origin or the whole plane. The literature provides various proofs of Yamabe's theorem: see [6] for example. A slightly weakened version of Yamabe's theorem is proved in [5, Appendix 4] in which every pair of points is assumed to be joined by a piecewise differentiable arc, and in [1] and [2] the theorem is proved for the special case that the group is in R^n. However none of the proofs is ``elementary'' in the sense of this paper even when applied to the most special of cases, the plane. This is unfortunate since an appreciation for the theorem in this ``intuitively obvious'' case should be at hand. By replacing `arcwise connected' with `connected', and making the additonal hypothesis that the subgroup must contain one special type of arc (a d-arc), it is our aim to provide an accessible proof.

    Note that if g: [a,b] \rightarrow R^2 then, by virtue of the Inverse Function Theorem, the set \{g(t) = (x(t),y(t)): a < t < b\} will be a d-arc as defined above provided that x(t) is a differentiable function with continuous derivative and x'(t_0) \neq 0 for some t_0 \in (a,b). This conclusion also holds if x(t) is replaced by y(t) in the statement.

    All subgroups of the plane will be assumed to be additive subgroups.

    We can now state a special case of Yamabe's theorem in the plane.


    A connected subgroup of the plane that contains a $d-$arc must be either a straight line through the origin or the whole plane.

    We point out that even in the case of the plane, Yamabe's theorem is not true if `arcwise connected' is replaced by `connected': see the example at the end of the paper. However, the additional hypothesis that the subgroup contains a d-arc is sufficient. We continue with two easy observations.

    Observation 1

    If $H$ is a subgroup of $R^2$, and it contains a line $L$ and a point $P(c,d)$ not on $L$, then $H$ contains the line $L'$ through $P$ and parallel to $L$.
    Figure 1

    Proof. Suppose that H contains the non-vertical line L = \{(x,mx+b): x \in R\}. Looking at Figure 1, we see that A-B+P = C which is an arbitray point on the line L'. Here A = (x,mx+b), B = (0,b) and C = (x+c, m(x+c) + (d-mc)). The proof is even simpler if L is a vertical line.

    Observation 2

    If $H$ is a subgroup of $R^2$, and it contains a line segment, then it contains the line through this line segment. In particular, if a subgroup of the plane contains an interval of the $X-$axis or the $Y-$axis then it must contain the entire $X-$axis or the entire $Y-$axis respectively.

    Proof. As in the proof of Observation 1, points on the given line segment S can be subtracted to obtain a line segment S' which is parallel to S through the origin. Then additions of points on S' include the entire line containing S' and hence, by Observation 1, the line containing S.

    We now come to the main result.


  • (1) {If $H$ is a (not necessarily connected) subgroup of $R^2$ and contains a nonlinear $d-$arc, then $H = R^2$.}
  • (2) {If $H$ is a connected subgroup of $R^2$ and contains a linear $d-$arc (a line segment) and a point not on the line containing this line segment, then $H = R^2$.}

    Outline of proof. To prove (1) we argue as follows. If the subgroup H contains a nonlinear d-arc, then we will use additivity of H to show that it must contain an interval of the Y-axis and so must contain the entire Y-axis. From this it is easy to shwo that H must also contain the entire X-axis and hence the whole plane. (Additivity again!)

    Proof of Proposition. Remark: The proof we give below is for the case that H contains a d-arc of the form \{(x,f(x)): a < x < c \}. A similar argument handles the case when the arc is of the form \{(f(x),x): a < x < c\}.

    Claim: If f is continuous then f is linear if and only if g(x,y) = f(x)+f(y)-2f((x+y)/2) = 0.

    Proof of claim: If f is linear then clearly g(x,y) = 0. On the other hand, suppose that g(x,y) = 0 for all x and y. In this case, the line segment joining the points A = (x,f(x)) and B = ((x+y)/2, f((x+y)/2)) = ((x+y)/2, (f(x)+f(y))/2) has the same slope as the line segment joining B and C = (y,f(y)). Therefore, for any values of x and y the three points A,B,C are collinear. Thus, by continuity of f, f is linear and the claim is shown.

    Note. If the function g above is a constant function, then substituting x=y into the equation for g shows that g must actually be equal to the zero function: that is g is constant if and only if g=0.

    We now return to the proof of the proposition.

    Figure 2

    (1) Choose real numbers x and y such that a < x,y < c. Then a < (x+y)/2 < c. Therefore the following three points lie in H (see Figure 2):
    A = (x,f(x)), B = (y,f(y)), C = ((x+y)/2, f((x+y)/2)).
    Since H is an additive group, the point D = A+B-2C = (0,g(x,y)) is also in H. Since f is non-linear, by the note above, g(x,y) = f(x)+f(y)-2f((x+y)/2) is not constant. Then, being a continuous function from R^2 to R that is not constant, the image of g must contain an interval (t_1,t_2), say, in R. This simply says that H contains the interval \{(0,r): t_1 < r < t_2\} of the Y-axis. It then follows from Observation 2 that H contains the entire Y-axis. Now for any x \in (a,c) we have (x,f(x)) \in H, and since H contains the entire Y-axis, H contains the point (x,f(x)) - (0,f(x)) = (x,0) for any x \in (a,c). Again, by Observation 2, H must now contain the X-axis. Since every point (s,t) in the plane is the sum of a point on the X-axis and a point on the Y-axis (because (s,t)=(s,0)+(0,t)), we have shown that H = R^2 if f is nonlinear.

    (2) Now suppose that H contains a linear d-arc and a point P which is not on the line L containing this arc. Let L' be the line through P and parallel to L. By Observations 1,2 respectively, H contains L' and L. Consider the subgroup S = H \cap Y-axis. It is a connected subgroup of the y-axis and so must either be zero (the origin) or the entire Y-axis. This follows from Observation 2 because a connected subset of the real line is either a point, an interval, or the whole line itself. Now L and L' have different y-intercepts and so S must be the entire Y-axis. This implies that H contains all lines parallel to L (again by Observation 1) and this is the entire plane.

    The following result is a special case of Yamabe's theorem in the plane.


    A connected subgroup of the plane which contains a $d-$arc must be either a straight line through the origin or the whole plane.

    Proof. If the d-arc is non-linear then by Proposition (1) the subgroup must be the entire plane. If the d-arc is linear and the subgroup contains a point not on the line containing this d-arc, then, by Proposition (2), the subgroup must be the entire plane. If the subgroup contains no points not on the line, then the subgroup is just a straight line through the origin.

    The following example (based on nontrivial results of F.B. Jones [3],[4]) shows that the hypothesis of `arcwise connected' in Yamabe's Theorem cannot be replaced by `connected'.

    Example 2. Although F.B. Jones does not specifically address Yamabe's Theorem, it is easy to make the connection from his work. Recall that a Hamel basis of R is a set of elements of R that is linearly independent over the rational numbers and is maximal with respect to this property. Jones uses a Hamel basis of R to define a function f:R \rightarrow R with the following properties.

  • (a) {$f(x+y) = f(x) + f(y)$}
  • (b) {$Gph(f)$ is a connected set.}
  • (c) {$Gph(f)$ is dense in the plane but contains no arc.}

    Here Gph(f) = \{(x,f(x)): x \in R\} denotes the graph of f.

    The set Gph(f) defined above is clearly a nontrivial, connected, additive subgroup of the plane. By (c) it is neither a straight line through the origin nor the entire plane. We refer the reader to [4] for the definition of f and the proof that it actually satisfies (a)-(c) above.


  • [1] {Hayashida, Tsuyoshi, `Arc-wise connected subgroup[s] of a vector group', Kodai Mathematical Seminar Reports, (1949) pp.16-17.}

  • [2] {Homma, Tatsuo, Minagawa and Takizo, `Vector-group[s] in real Euclidean space', Kodai Mathematical Seminar Reports, (1949) pp.19-20.}

  • [3] {Jones, F.B., `Measure and other properties of a Hamel basis', Bull. Amer. Math. Soc. v.48 (1942) pp.472-481.}

  • [4] {Jones, F.B., `Connected and disconnected plane sets and the functional equation $f(x)+f(y)=f(x+y)$', Bull. Amer. Math. Soc. v.48 (1942) pp.115-120.}

  • [5] {Kobayashi, Shoshichi and Nomizu, Katsumi, Foundations of Differential Geometry, vol. 1. New York: Interscience Publishers, 1963.}

  • [6] {Yamabe, Hidehiko, `On an arcwise connected subgroup of a Lie group', Osaka Mathematical Journal v.2, no. 1 Mar. (1950) pp.13-14.}

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