(c) If K is a subgroup of G, then p(K) is a subgroup of H. Given: K < G (d) If K' is a subgroup of H, then p-'(K') is a subgroup of G.

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I did a,b.......need help with c &d for abstract algebra.

 

1. Let \( \varphi: G \rightarrow H \) be a group homomorphism. Show:

(a) \( \varphi(1_G) = 1_H \)

Let \( \varphi: G \rightarrow H \)

\[ \varphi(1_G) = 1_H, \] where \( 1_G \) and \( 1_H \) are the identities of \( G \) and \( H \) respectively.

\[ \varphi(1_G) = \varphi(1_G 1_G) \]

\[ = \varphi(1_G) \varphi(1_G) \quad (\varphi \text{ is a homomorphism}) \]

\[ \varphi(1_G)^{-1} \varphi(1_G) = \varphi(1_G)^{-1} \varphi(1_G) \varphi(1_G) \]

\[ 1_H = 1_H \varphi(1_G) \]

\[ 1_H = \varphi(1_G) \]

\(\therefore \varphi(1_G) = 1_H \) \[  \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (*)\]

(b) \( \varphi(a^{-1}) = [\varphi(a)]^{-1}, \forall a \in G; \)

\[ \varphi(g^{-1}) = \varphi(g)^{-1} \forall g \in G \]

\[ \varphi(1_G) = \varphi(gg^{-1}) = \varphi(g) \varphi(g^{-1}) \]

Since \( \varphi(1_G) = 1_H \) \[  \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (*)\]

\(\Rightarrow 1_H = \varphi(g) \varphi(g^{-1}) \)

\(\Rightarrow \varphi(g^{-1}) 1_H = \varphi(g) \varphi(g^{-1}) \varphi(g^{-1}) \)

\(\Rightarrow \varphi(g^{-1}) = \
Transcribed Image Text:1. Let \( \varphi: G \rightarrow H \) be a group homomorphism. Show: (a) \( \varphi(1_G) = 1_H \) Let \( \varphi: G \rightarrow H \) \[ \varphi(1_G) = 1_H, \] where \( 1_G \) and \( 1_H \) are the identities of \( G \) and \( H \) respectively. \[ \varphi(1_G) = \varphi(1_G 1_G) \] \[ = \varphi(1_G) \varphi(1_G) \quad (\varphi \text{ is a homomorphism}) \] \[ \varphi(1_G)^{-1} \varphi(1_G) = \varphi(1_G)^{-1} \varphi(1_G) \varphi(1_G) \] \[ 1_H = 1_H \varphi(1_G) \] \[ 1_H = \varphi(1_G) \] \(\therefore \varphi(1_G) = 1_H \) \[ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (*)\] (b) \( \varphi(a^{-1}) = [\varphi(a)]^{-1}, \forall a \in G; \) \[ \varphi(g^{-1}) = \varphi(g)^{-1} \forall g \in G \] \[ \varphi(1_G) = \varphi(gg^{-1}) = \varphi(g) \varphi(g^{-1}) \] Since \( \varphi(1_G) = 1_H \) \[ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (*)\] \(\Rightarrow 1_H = \varphi(g) \varphi(g^{-1}) \) \(\Rightarrow \varphi(g^{-1}) 1_H = \varphi(g) \varphi(g^{-1}) \varphi(g^{-1}) \) \(\Rightarrow \varphi(g^{-1}) = \
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