# Let $(A, \star)$, and $(B, \diamond)$ be groups and let $A \times B$ be their direct product. Prove that $(1, 1)$ is the identity of $A \times B$

Let $(A, \star)$, and $(B, \diamond)$ be groups and let $A \times B$ be their direct product. Prove that $(1, 1)$ is the identity of $A \times B$

Recall if $(A, \star)$, and $(B, \diamond)$ are groups, their direct product is the group $D = (A \times B,\ \bullet)$, such that $(a_1, b_1) \bullet(a_2, b_2) = (a_1 \star a_2, b_1 \diamond b_2)$.

Now I can't see why $(1, 1)$ need be the identity of $A \times B$. An easy counter-example is if we take the two groups to be $\mathbb{R}$ under addition, then $D = (\mathbb{R^2}, \bullet)$, and $(0, 0)$ should be the identity.

More generally, I'm sure the identity of $A \times B$ would depend on the binary operations on $A$ and $B$ respectively, and since these binary operations $\star$ and $\diamond$ are completely arbitrary, I don't see how we could determine the identity for $A \times B$

Is there an error in the original question? (this was taken from an exercise in Abstract Algebra by Dummit and Foote)

• In this context, $1$ is simply the name for the identity element of the group, whatever that may be. Commented Dec 28, 2016 at 5:36
• @florence As opposed to $1 \in \mathbb{Z}$ or $\mathbb{R}$ or $\mathbb{Q}$ Commented Dec 28, 2016 at 5:37
• Right, exactly. Commented Dec 28, 2016 at 5:41

You are correct. What D&F means to say is:

Let $A$ and $B$ be groups, and let $e_{_A}$ be the identity element in $A$ and $e_{_B}$ the identity element in $B$. Then the element $(e_{_A}, e_{_B})$ is the identity element in $A \times B$.

If the group / operation is unspecified, it is sometimes the case that people will choose "$1$" to be the default notation for identity element.