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<p>(We say an object $a$ is an identity of $*$ if $a*b=b$, for all $b$ in the set that's being considered.)</p>
<p>These operations are only examples - can you find other possible answers?</p>
<p>$\mathbf{X_1}$: $* = \times$ (mod 2), $\dagger = +$ (mod 2).</p>
<p>Here, $0$ is the additive identity, and $1$ is the multiplicative identity.</p>
<p>$\mathbf{X_6}$: $* = \times$ (mod 7), $\dagger = +$ (mod 7). You should check these operations are closed, i.e. $a*b$ and $a\dagger b$ are in $X_7$ for all $a$ and $b$. Again, $0$ is the additive identity, and $1$ is the multiplicative identity.</p>
<p>The operations don't necessarily need to have a sensible interpretation, however: here's a perfectly good $*$ operation (with identity $4$) defined in terms of a multiplication table:</p>
<table id="multiplicationtable">
<thead>
<tr>
<th scope="row">$*$</th>
<th scope="col">0</th>
<th scope="col">1</th>
<th scope="col">2</th>
<th scope="col">3</th>
<th scope="col">4</th>
<th scope="col">5</th>
<th scope="col">6</th>
</tr>
</thead>
<tbody>
<tr>
<th scope="row">0</th>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>0</td>
<td>1</td>
<td>1</td>
</tr>
<tr>
<th scope="row">1</th>
<td>1</td>
<td>2</td>
<td>0</td>
<td>4</td>
<td>1</td>
<td>3</td>
<td>3</td>
</tr>
<tr>
<th scope="row">2</th>
<td>1</td>
<td>0</td>
<td>5</td>
<td>2</td>
<td>2</td>
<td>3</td>
<td>0</td>
</tr>
<tr>
<th scope="row">3</th>
<td>1</td>
<td>4</td>
<td>2</td>
<td>6</td>
<td>3</td>
<td>2</td>
<td>3</td>
</tr>
<tr>
<th scope="row">4</th>
<td>0</td>
<td>1</td>
<td>2</td>
<td>3</td>
<td>4</td>
<td>5</td>
<td>6</td>
</tr>
<tr>
<th scope="row">5</th>
<td>1</td>
<td>3</td>
<td>3</td>
<td>2</td>
<td>5</td>
<td>6</td>
<td>6</td>
</tr>
<tr>
<th scope="row">6</th>
<td>1</td>
<td>3</td>
<td>0</td>
<td>3</td>
<td>6</td>
<td>6</td>
<td>2</td>
</tr>
</tbody>
</table>
<p><strong>Natural numbers </strong>$\mathbb{N}$: We can set $* = \times$ and $\dagger = +$. In this case, $1*b=b$ for all b, but there is no identity element for $\dagger$, as $0$ is not in the set.</p>
<p>$\mathbf{2}$ <strong>by</strong> $\mathbf{2}$ <strong>matrices</strong>: Consider $A=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ and $A=\left(\begin{smallmatrix}e&f\\g&h\end{smallmatrix}\right)$. Suppose all the matrix entries are integers. We could define operations: $A*B=\left(\begin{smallmatrix}a+e&b+f\\c+g&d+h\end{smallmatrix}\right)$ (i.e. component
addition). All of these components are integers, so this operation is closed. This has the identity:$\left(\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right)$.</p>
<p>$A\dagger B=\left(\begin{smallmatrix}a\times e&b\times f\\c\times g &d\times h\end{smallmatrix}\right)$ (component multiplication). Again, this operation is closed. This has an identity: $\left(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\right)$</p>
<p><strong>Words:</strong> Given two strings of length $n$, $l = l_1l_2l_3\dots$ and $m=m_1m_2m_3\dots$ say, each comprising of two symbols $a$ and $b$, we could choose operations that act independently on $l_1$ and $m_1$, on $l_2$ and $m_2$, $\dots$ Suppose the operation $*$ returned $a$ in the ith position if $l_i$ and $m_i$ were the same, and $b$ if they were different. Also, suppose the
operation $\dagger$ returned $b$ if $l_i$ and $m_i$ were the same, and $a$ if they were different. The identity of $*$ is the string $aaaaaa\dots$ and the identity of $\dagger$ is the string $bbbbb\dots$</p>
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