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Math and science::Algebra::Aluffi

Subgroups. Propositions.

The proofs of the following are on the reverse side. Can you remember them?

Image of subgroup is a subgroup in \( \cat{Grp} \)

Let \( \varphi : G \to H \) be a group homomorphism, and let \( G' \) be a subgroup of \( G \). Then the image of \( G' \) through \( \varphi \) is a subgroup of \( H \).

Let \( \varphi : G \to H \) be a group homomorphism, and let \( H' \) be a subgroup of \( H \). Then \( \inv{\varphi}(H) \) is a subgroup of \( G \).

The proof has two steps, one for each of the requisite conditions of a subgroup.

Let \( H_1 \) and \( H_2 \) be two subgroups of \( G \). Then \( H_1 \cap H_2 \) is a subgroup of \( G \).


2.

Proof. Let \( G' = \inv{\varphi}(H) \). We must show that \( G' \) satisfies the two subgroup conditions.

  • Let \( a, b \in G' \), then consider \( m_{G}(a, b) \). Denote \( g^* = m_{G}(a, b) \) and \( h^* = \varphi( m_G(a, b)) \). As \( \varphi \) is a group homomorphism, \( h^* = m_H(\varphi(a), \varphi(b)) \). \( H' \) is a subgroup and thus closed with respect to \( m_H \), so \( h^* \in H' \). This means that \( g^* \) is in \( G' \). As \( a \) and \( b \) were arbitrary, this shows that \( G' \) is closed under \( m_G \).
  • Let \( a \in G' \). Being a group homomorphism, \( \varphi \) preserves inverses, so \( \inv{a} \) must map to \( \inv{\varphi(a)} \). This element is in \( H' \), as \( \varphi(a) \in H' \) and \( H' \) must contain its inverses.

3.

Essence. Elements in the intersection must have their inverses in both subsets, and any output of a product must be in both subsets too. So the intersection cotains the inverses and is closed under the group operation. Of course, the identity is in both subsets.

Subgroups are recapped below.

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 \( m_G \).

Source

Aluffi, p80