Here we will be looking at direct products of groups. In particular, we will look at how the taking of a direct product or decomposition preserves commutativity.

**Preservation of Commutativity**

Let \(A\) and \(B\) be be groups. Then \(A \times B\) is commutative if and only if \(A\) and \(B\) are both commutative.

**\(A\) and \(B\) commutative**

The given theorem is an if and only if statement, so we will need to give an if then proof in two directions. We will start by showing that if \(A\) and \(B\) are commutative, then \(A \times B\) is commutative.

The first thing we will need to recall is that a group \(G\) is commutative if, for all \(a,b \in G\), we have that \(ab=ba\).

Next, we need that \(A\times B=\left\{(a,b):a \in A, b \in B\right\}\) and \((a,b)(c,d)=(ac,bd)\) for all \(a,c \in A\) and \(b,d \in B\).

With these stated, we now need to show that for every \((a,b),(c,d) \in A \times B\), we have that \((a,b)(c,d)=(c,d)(a,b)\). However, if we note that \((a,b)(c,d)=(ac,bd)=(ca,db)=(c,d)(b,a)\) since \(A\) and \(B\) are commutative, we have our proof.

**Proof**

Let \(A\) and \(B\) be commutative groups. Then, let \((a,b), (c,d) \in A \times B\). We then have that

\begin{align*}

(a,b)(c,d)&=(ac,bd) \\

&=(ca, db),

\end{align*}

since \(A\) and \(B\) are commutative. Then

\begin{align*}

(ca,db)&=(c,d)(a,b).

\end{align*}

Hence \((a,b)(c,d)=(c,d)(a,b)\) for all \((a,b),(c,d) \in A \times B\), so \(A \times B\) is commutative.

**\(A \times B\) is commutative**

To show the other direction, we will assume that \(A \times B\) is commutative and we must show that both \(A\) and \(B\) are commutative. Therefore, let \(a,c \in A\) and \(b,d \in B\). We then need to show that \(ac=ca\) and \(bd=db\). In order to do this, we must recall that \((x,y)=(w,z)\) if and only if \(x=w\) and \(y=z\). Because of this, we see that

\begin{align*}

(ac,bd)&=(a,b)(c,d) \\

&=(c,d)(a,b) \\

&=(ca,db),

\end{align*}

with the second line following from \(A\times B\) being commutative. Since the two ordered pairs are equal, this implies the entries are equal, so we get \(ac=ca\) and \(bd=db\).

**Proof**

Let \(A\) and \(B\) be groups with \(A \times B\) commutative. Then, let \(a,c \in A\) and \(b,d \in B\). We then note that

\begin{align*}

(ac,bd)&=(a,b)(c,d) \\

&=(c,d)(a,b) \\

&=(ca,db).

\end{align*}

Since the entries of the ordered pair are equal, we have \(ac=ca\) and \(bd=db\). Hence \(A\) and \(B\) are commutative.

**Conclusion**

We have now shown that commutativity is preserved when taking direct products and when looking at components of direct products. While it is not immediately clear, it is indeed true that any equational identity would be preserved under the taking of direct product or the decomposition (this a portion of the HSP theorem). Even though we may not be able to show the general case in this class, we could work, in a similar way, through any given identity.

I hope you this helps you with your study of algebra. If it did, make sure to check out the other posts at Algebraic Structures.

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