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Math and science::Algebra::Aluffi

Subgroups. Definition.

Subgroup

Let \( (G, m_G ) \) be a group. A group \( (H, m_H) \) is a subgroup of \( G \) iff both of the following conditions hold:

  1. \( H \subseteq G \)
  2. The inclusion function \( i: H \to G \) forms a group homomorphism.

An alternative form of the subgroup condition is the following pair of conditions:

  1. \( \forall g \in G, \; g \in H \implies g^{-1} \in H \).
  2. \( \forall g_1, g_2 \in G, \; g_1, g_2 \in H \implies m_G(g_1, g_2) \in H \).

In words, these two conditions say:

  1. For all elements of \( H \), their inverses are also in \( H \) also..
  2. \( H \) is closed with respect to the operation \( m_G \).

Aluffi has a condensed formula that combines these two conditions into one. Can you remember it?


Alternative definition.

A more constructive definition:

Subgroup. Definition.

Let \( (G, m_G ) \) be a group. A subgroup of \( G \) is a group \( (H, m_H ) \) whose underlying set \( H \) is a subset of \( G \) and whose operation \( m_H \) is a valid function \( m_H : H \times H \to H \) defined by inheriting its mapping from \( m_G \).

There is some heavy lifting being done by the phrases "is a valid function" and "inheriting its mapping":

  • "Is a valid function" says the function \( m_G \) must be closed for elements of \( H \).
  • "Inherits its mapping" says that:
    \[ \begin{align} & \forall h_1, h_2 \in H, && \\ & \quad m_H(h_1, h_2) \text{ is the element } h_3 \text{ such that } i(h_3) = m_G(i(h_1), i(h_2)), \\ &\text{where } i : H \to G \text{ is the inclusion. } \end{align} \]

This requirement of \( m_H \) is equivalent to stating that the inclusion function \( i : H \to G \) forms a group homomorphism.

The two definitions are equivalent, but this should be proved.

Condensed subgroup condition

Aluffi packages the two subgroup conditions into a tight statement:

\[ \forall a, b \in H, \; ab^{-1} \in H \]

I don't like this statement for a number of reasons. Firstly, it's not clear what the operation is. Also, It is using the presence of the symbol \( b^{-1} \) to replace the statement "\( b \) must have an inverse", which seems like an abuse: \( b \) is the symbol introduced by the \( \forall \), not \( b^{-1} \). Under charitable circumstances one could say that \( x^{-1} \) is syntax for \( \operatorname{inv}(x) \), where \( \operatorname{inv} \) is a function either \( \operatorname{inv} G \to G \) or \( \operatorname{inv} H \to H \). But even in this case, the existence of this function and the fact that it maps to inverses correctly is not stated at all.

Source

Aluffi, p79