# Group Direct Product/Examples/C2 x C2/Subgroups

## Subgroups of Group Direct Product of $C_2 \times C_2$

Let us represent $C_2$ as the group $\struct {\set {1, -1}, \times}$:

- $\begin {array} {r|rr} \struct {\set {1, -1} , \times} & 1 & -1 \\ \hline 1 & 1 & -1 \\ -1 & -1 & 1 \\ \end{array}$

The subsets of $C_2 \times C_2$ which form its subgroups are:

- $\set {\tuple {1, 1} }$
- $\set {\tuple {1, 1}, \tuple {1, -1} }$
- $\set {\tuple {1, 1}, \tuple {-1, 1} }$
- $\set {\tuple {1, 1}, \tuple {-1, -1} }$
- $\set {\tuple {1, 1}, \tuple {1, -1}, \tuple {-1, 1} , \tuple {-1, -1} }$ (that is, $C_2 \times C_2$ itself)

## Proof

From Trivial Subgroup is Subgroup and Group is Subgroup of Itself we have that $\set {\tuple {1, 1} }$ and $C_2 \times C_2$ are two of the subgroups

From Lagrange's Theorem (Group Theory), any subgroup of $C_2 \times C_2$ contains $1$, $2$ or $4$ elements.

There is only one subgroup for each of $1$ and $4$, both of which have been covered.

Thus all other subgroups can contain only $2$ elements.

Every subgroup of $C_2 \times C_2$ contains its identity $\set {\tuple {1, 1} }$.

So a subgroup of $C_2 \times C_2$ contains $\set {\tuple {1, 1} }$ and one other element.

Each of the other elements is of order $2$.

So, along with $\set {\tuple {1, 1} }$, each one forms a subgroup of $C_2 \times C_2$.

$\blacksquare$

## Sources

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $5$: Cosets and Lagrange's Theorem: Example $5.11$