Question

Direct Product Recognition Theorem: Let G be a group, let H and K be subgroups of G. Show that the following are equivalent: (i) The function $H \times K \to G$, $(h, k) \mapsto hk$ is an isomorphism. (ii) For every $h \in H$ and $k \in K$, one has $hk = kh$ and for every $g \in G$, there exist unique elements $h \in H$ and $k \in K$ such that $g = hk$. (iii) $H \cap K = 1$, $HK = G$ and for every $h \in H$ and $k \in K$ one has $hk = kh$. (iv) $H \trianglelefteq G$, $K \trianglelefteq G$, $H \cap K = 1$ and $HK = G$.

          Direct Product Recognition Theorem: Let G be a group, let H and K be
subgroups of G. Show that the following are equivalent:
(i) The function $H \times K \to G$, $(h, k) \mapsto hk$ is an isomorphism.
(ii) For every $h \in H$ and $k \in K$, one has $hk = kh$ and for every $g \in G$, there exist unique elements
$h \in H$ and $k \in K$ such that $g = hk$.
(iii) $H \cap K = 1$, $HK = G$ and for every $h \in H$ and $k \in K$ one has $hk = kh$.
(iv) $H \trianglelefteq G$, $K \trianglelefteq G$, $H \cap K = 1$ and $HK = G$.

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Direct Product Recognition Theorem: Let G be a group, let H and K be
subgroups of G. Show that the following are equivalent:
(i) The function H × K → G, (h, k) ↦ hk is an isomorphism.
(ii) For every h ∈ H and k ∈ K, one has hk = kh and for every g ∈ G, there exist unique elements
h ∈ H and k ∈ K such that g = hk.
(iii) H ∩ K = 1, HK = G and for every h ∈ H and k ∈ K one has hk = kh.
(iv) H ⊴ G, K ⊴ G, H ∩ K = 1 and HK = G.

Added by Isabel C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Suman K

06/28/2022

Step 1

Assume that the function H x K -> G, (h,k) |-> hk is an isomorphism. This means that the function is bijective and preserves the group operation. So for every h in H and k in K, we have hk = kh, and for every g in G, there exist unique elements h in H and k in K Show more…

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Direct Product Recognition Theorem: Let G be a group, let H and K be subgroups of G. Show that the following are equivalent: (i) The function H x K -> G, (h,k) |-> hk is an isomorphism. (ii) For every h in H and k in K, one has hk = kh and for every g in G, there exist unique elements h in H and k in K such that g = hk. (iii) H ∩ K = 1, HK = G and for every h in H and k in K one has hk = kh. (iv) H ⊴ G, K ⊴ G, H ∩ K = 1 and HK = G.

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