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Simon
Simon
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As an example, consider the graphs $G$ and $G'$ on 4 vertices, labelled 1, 2, 3 and 4, where $G$ has edge set $\{\{1,2\},\{1,3\},\{2,3\},\{3,4\}\}$ and $G'$ has edge set $\{\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}$. Sketch both of these graphs !

Then the permutation $\alpha = (1, 2, 3, 4)$ (in disjoint cycle notation) $\alpha = (1, 2, 3, 4)$ is an isomorphism from $G$ to $G'$. Why ? Because, applying $\alpha$ to the vertex labels in the edge set of $G$ we obtain the edge set of $G'$. Check this ! Since an isomorphism from $G$ to $G'$ exists, $G$ and $G'$ are isomorphic. However, since their edge sets are different, $G$ and $G'$ are not equal.

The permutation $\beta = (1, 2)$ is an isomorphism from $G$ to $G$ itself, that is, an automorphism of $G$, also known as a symmetry of $G$. Check this, as above !

As an example, consider the graphs $G$ and $G'$ on 4 vertices, labelled 1, 2, 3 and 4, where $G$ has edge set $\{\{1,2\},\{1,3\},\{2,3\},\{3,4\}\}$ and $G'$ has edge set $\{\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}$. Sketch both these graphs !

Then the permutation (in disjoint cycle notation) $\alpha = (1, 2, 3, 4)$ is an isomorphism from $G$ to $G'$. Why ? Because, applying $\alpha$ to the vertex labels in the edge set of $G$ we obtain the edge set of $G'$. Check this ! Since an isomorphism from $G$ to $G'$ exists, $G$ and $G'$ are isomorphic. However, since their edge sets are different, $G$ and $G'$ are not equal.

The permutation $\beta = (1, 2)$ is an isomorphism from $G$ to $G$ itself, that is, an automorphism of $G$, also known as a symmetry of $G$. Check this, as above !

As an example, consider the graphs $G$ and $G'$ on 4 vertices, labelled 1, 2, 3 and 4, where $G$ has edge set $\{\{1,2\},\{1,3\},\{2,3\},\{3,4\}\}$ and $G'$ has edge set $\{\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}$. Sketch both of these graphs !

Then the permutation $\alpha = (1, 2, 3, 4)$ (in disjoint cycle notation) is an isomorphism from $G$ to $G'$. Why ? Because, applying $\alpha$ to the vertex labels in the edge set of $G$ we obtain the edge set of $G'$. Check this ! Since an isomorphism from $G$ to $G'$ exists, $G$ and $G'$ are isomorphic. However, since their edge sets are different, $G$ and $G'$ are not equal.

The permutation $\beta = (1, 2)$ is an isomorphism from $G$ to $G$ itself, that is, an automorphism of $G$, also known as a symmetry of $G$. Check this, as above !

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Simon
Simon
  • 2.2k
  • 18
  • 25

As an example, consider the graphs $G$ and $G'$ on 4 vertices, labelled 1, 2, 3 and 4, where $G$ has edge set $\{\{1,2\},\{1,3\},\{2,3\},\{3,4\}\}$ and $G'$ has edge set $\{\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}$. Sketch both these graphs !

Then the permutation (in disjoint cycle notation) $\alpha = (1, 2, 3, 4)$ is an isomorphism from $G$ to $G'$. Why ? Because, applying $\alpha$ to the vertex labels in the edge set of $G$ we obtain the edge set of $E$$G'$. Check this ! Since an isomorphism from $G$ to $G'$ exists, $G$ and $G'$ are isomorphic. However, since their edge sets are different, $G$ and $G'$ are not equal.

The permutation $\beta = (1, 2)$ is an isomorphism from $G$ to $G$ itself, that is, an automorphism of $G$, also known as a symmetry of $G$. Check this, as above !

As an example, consider the graphs $G$ and $G'$ on 4 vertices, labelled 1, 2, 3 and 4, where $G$ has edge set $\{\{1,2\},\{1,3\},\{2,3\},\{3,4\}\}$ and $G'$ has edge set $\{\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}$. Sketch both these graphs !

Then the permutation (in disjoint cycle notation) $\alpha = (1, 2, 3, 4)$ is an isomorphism from $G$ to $G'$. Why ? Because, applying $\alpha$ to the vertex labels in the edge set of $G$ we obtain the edge set of $E$. Check this ! Since an isomorphism from $G$ to $G'$ exists, $G$ and $G'$ are isomorphic. However, since their edge sets are different, $G$ and $G'$ are not equal.

The permutation $\beta = (1, 2)$ is an isomorphism from $G$ to $G$ itself, that is, an automorphism of $G$, also known as a symmetry of $G$. Check this, as above !

As an example, consider the graphs $G$ and $G'$ on 4 vertices, labelled 1, 2, 3 and 4, where $G$ has edge set $\{\{1,2\},\{1,3\},\{2,3\},\{3,4\}\}$ and $G'$ has edge set $\{\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}$. Sketch both these graphs !

Then the permutation (in disjoint cycle notation) $\alpha = (1, 2, 3, 4)$ is an isomorphism from $G$ to $G'$. Why ? Because, applying $\alpha$ to the vertex labels in the edge set of $G$ we obtain the edge set of $G'$. Check this ! Since an isomorphism from $G$ to $G'$ exists, $G$ and $G'$ are isomorphic. However, since their edge sets are different, $G$ and $G'$ are not equal.

The permutation $\beta = (1, 2)$ is an isomorphism from $G$ to $G$ itself, that is, an automorphism of $G$, also known as a symmetry of $G$. Check this, as above !

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Simon
Simon
  • 2.2k
  • 18
  • 25

As an example, consider the graphs $G$ and $G'$ on 4 vertices, labelled 1, 2, 3 and 4, where $G$ has edge set $\{\{1,2\},\{1,3\},\{2,3\},\{3,4\}\}$ and $G'$ has edge set $\{\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}$. Sketch both these graphs !

Then the permutation (in disjoint cycle notation) $\alpha = (1, 2, 3, 4)$ is an isomorphism from $G$ to $G'$. Why ? Because, applying $\alpha$ to the vertex labels in the edge set of $G$ we obtain the edge set of $E$. Check this ! Since an isomorphism from $G$ to $G'$ exists, $G$ and $G'$ are isomorphic. However, since their edge sets are different, $G$ and $G'$ are not equal.

The permutation $\beta = (1, 2)$ is an isomorphism from $G$ to $G$ itself, that is, an automorphism of $G$, also known as a symmetry of $G$. Check this, as above !