# Why isn't this a subgroup?

The addition of integers $$\mod N$$ generates a group. For example, under addition mod 5 the set $$\{0, 1, 2, 3, 4\}$$ forms a group: $$2 + 1= 3, 3 + 2 = 0, 4 + 3 = 2,$$ and so on. The composition of the group elements is defined by $$j * k = j + k \mod 5$$. The identity element $$I$$ is denoted by $$0$$. The inverse of $$2$$, for example, is $$3$$, of $$4$$ is $$1$$, and so on. The group is clearly abelian.

This example is from A. Zee Group Theory in a Nutshell for Physicists

I want to construct a subgroup.So first i take the identity element $$0$$. Now i have a trivial subgroup. Then i add another element $$1$$ also i need its inverse that is $$4$$. The inverse of $$4$$ is $$1$$.

$$\{0, 1, 4\}$$ is a group under addition mod5 , right ? It is closed, it has an identity, every element has an inverse, it is associative.

But it cannot be a group ( subgroup ) due to the Lagrange theorem because it has $$3$$ elements and $$3$$ does not divide $$5$$.

It is not closed. $$1+1=2,$$ which is not in $$\{0,1,4\}.$$

It's not closed because $$1+1$$ isn't in it.

1+1=2 and 4+4=3 is not in your set. So it is not a semigroup. Ok, now you can restrict your operation as, you will not alow a*a, i.e, if you say that the operation between same element is not allowed, i.e., 1+1 and 4+4 is not defined then {0,1,4} is closed. But it is not so much interesting.

The smallest subgroup containing $$1$$, should also contains $$1+1, 1+1+1, 1+1+1+1 \dots$$ and also their inverses.

The group operation ( = group law ) $$\color{red}{+}$$ here is $$a\color{red}{+}b = (a+b)\bmod 5$$ and $$+$$ is the normal addition of integers. The set $$G=\{0,1,2,3,4\}$$ together with $$\color{red}{+}$$ build a group $$(G,\color{red}{+})$$.

We can operate on all elements in the set $$G=\{0,1,2,3,4\}$$ and get the following

$$0\color{red}{+}0 = (0+0)\bmod 5 = 0 = 0\color{red}{+}0$$

$$0\color{red}{+}1 = (0+1)\bmod 5 = 1$$

$$0\color{red}{+}2 = (0+2)\bmod 5 = 2$$

$$0\color{red}{+}3 = (0+3)\bmod 5 = 3$$

$$0\color{red}{+}4 = (0+4)\bmod 5 = 4$$

$$1\color{red}{+}0 = (1+0)\bmod 5 = 1 = 0\color{red}{+}1$$

$$1\color{red}{+}1 = (1+1)\bmod 5 = 2$$

$$1\color{red}{+}2 = (1+2)\bmod 5 = 3$$

$$1\color{red}{+}3 = (1+3)\bmod 5 = 4$$

$$1\color{red}{+}4 = (1+4)\bmod 5 = 0$$

$$2\color{red}{+}0 = (2+0)\bmod 5 = 2 = 0\color{red}{+}2$$

$$2\color{red}{+}1 = (2+1)\bmod 5 = 3 = 1\color{red}{+}2$$

$$2\color{red}{+}2 = (2+2)\bmod 5 = 4$$

$$2\color{red}{+}3 = (2+3)\bmod 5 = 0$$

$$2\color{red}{+}4 = (2+4)\bmod 5 = 1$$

$$3\color{red}{+}0 = (3+0)\bmod 5 = 3 = 0\color{red}{+}3$$

$$3\color{red}{+}1 = (3+1)\bmod 5 = 4 = 1\color{red}{+}3$$

$$3\color{red}{+}2 = (3+2)\bmod 5 = 0 = 2\color{red}{+}3$$

$$3\color{red}{+}3 = (3+3)\bmod 5 = 1$$

$$3\color{red}{+}4 = (3+4)\bmod 5 = 2$$

$$4\color{red}{+}0 = (4+0)\bmod 5 = 4 = 0\color{red}{+}4$$

$$4\color{red}{+}1 = (4+1)\bmod 5 = 0 = 1\color{red}{+}4$$

$$4\color{red}{+}2 = (4+2)\bmod 5 = 1 = 2\color{red}{+}4$$

$$4\color{red}{+}3 = (4+3)\bmod 5 = 2 = 3\color{red}{+}4$$

$$4\color{red}{+}4 = (4+4)\bmod 5 = 3$$

Or in a so called Cayley's table:

(1) neutral or identity element of $$G$$ with respect to the operation $$\color{red}{+}$$ is $$0 \in G$$

$$\forall a \in G:~ ~a\color{red}{+}0=a=0\color{red}{+}a.$$

Because we have the following (see first line in each block of calculation or first row and first column in the cayley's table):

$$0\color{red}{+}0 = 0$$, $$1\color{red}{+}0 = 1 = 0\color{red}{+}1$$ , $$2\color{red}{+}0 = 2 = 0\color{red}{+}2$$ , $$3\color{red}{+}0 = 3 = 0\color{red}{+}3$$ and $$4\color{red}{+}0 = 4 = 0\color{red}{+}4$$

(2) Inverse element of each element $$\in G$$ has to be also $$\in G$$

$$\forall a \in \{0,1,2,3,4\}:~ \exists b \in \{0,1,2,3,4\}:~ ~a\color{red}{+}b=0=b\color{red}{+}a$$

Because we have the following (see the lines in each calculation block, where there's "a $$0$$ in the middle", see also "the $$0$$ diagonale" in the cayley's table):

$$0\color{red}{+}0 = 0= 0\color{red}{+}0, 3\color{red}{+}2 = 0= 2\color{red}{+}3$$ and $$1\color{red}{+}4 = 0= 4\color{red}{+}1$$

(3) Closure with respect to the group operation/law $$\color{red}{+}$$

The group operation/law is a function $$\color{red}{+}: G \times G \rightarrow G, x \color{red}{+} y = (x+y) \bmod 5$$, which means that $$G$$ has to be closed under it's operation. So, $$\color{red}{+}$$ operates (a binary operator) on two arguments $$x \in G$$ and $$y \in G$$. Then it gives an element $$x\color{red}{+}y$$ back, which has to be again in $$G$$.

(4) Associativity with respect to the group operation/law $$\color{red}{+}$$

Assume $$a,b,c \in G$$, then

$$(a\color{red}{+}b)\color{red}{+}c=((a+b)\bmod 5 )\color{red}{+}c=(((a+b)\bmod 5)+c)\bmod 5$$

and

$$a\color{red}{+}(b\color{red}{+}c)=a\color{red}{+}((b+c) \bmod 5 )=(a+((b+c)\bmod 5))\bmod 5$$

So, if $$(a\color{red}{+}b)\color{red}{+}c=a\color{red}{+}(b\color{red}{+}c)$$ or if $$G$$ is associative, then $$(((a+b)\bmod 5)+c)\bmod 5 =(a+((b+c)\bmod 5))\bmod 5$$ has to be true. This $$(a+b)\bmod 5+c\equiv a+(b+c)\bmod 5 ~ ~(\bmod 5)$$ can be shown using modular arithmetic.

Or it can also be shown using the light's associativity test using the caley's table.

We can also see, that this group is commutative ($$a+b=b+a$$, symmetric cayley's table), which means that this group is a special group, it's an abelian group.

A sub group $$H < G$$ of a group $$G$$ has to satisfy all the four group properties or has to be a group for itself with respect to the same group operation $$\color{red}{+}$$.

Assume $$H = \{0,1,4\}$$. Then $$4 \color{red}{+} 4 = (4+4) \bmod 5 = 3 \not\in H$$ and $$1 \color{red}{+} 1 = (1+1) \bmod 5 = 2 \not\in H$$. $$H$$ is therefore not closed with respect to $$\color{red}{+}$$.

Maybe if we look close enough in the cayley's table, we might find a subgroup? But I don't think so, according to the Lagrange's theorem and due to the fact that $$| H | = 3 \nmid 5 = | G |$$.

For example here is a picture about other abelian groups from the wikipedia article about subgroups. The operation here is $$\color{red}{+}: G \times G \rightarrow G, x\color{red}{+}y = (x+y) \bmod 8$$

If $$G$$ is a (n abelian) groub, then $$H$$ is a (n abelian) subgroub.