Abstract algebra (direct product and isomorphism)

If G is a finite group and N_1, N_2 are two normal subgroups
of G such that |N_1||N_2|=|G| and N_1∩N_2=(e), then prove
that G ≃ N_1×N_2.

First, I claim that G = N₁N₂.
That N₁N₂ is a subgroup of G is clear. Hence, |N₁N₂| divides |G| by Lagrange's Theorem.
The opposite containment follows from |G| = |N₁| |N₂|.
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Next, we show that any g in G can be written uniquely in the form g = hk, where h ∈ N₁, k ∈ N₂.
Since G = N₁N₂, g = hk for some h ∈ N₁, k ∈ N₂.
Now, suppose that g = h'k' for some h' ∈ N₁, k' ∈ N₂.
So, hk = h'k' ==> (h')⁻¹h = k' k⁻¹.
However, since N₁ ∩ N₂ = {e}, we conclude that (h')⁻¹h = e = k' k⁻¹, and thus h = h', k = k'.
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Before displaying the isomorphism, note that N₁ and N₂ commute element-wise.
If h ∈ N₁, k ∈ N₂, then h⁻¹ k⁻¹ hk = (h⁻¹ k⁻¹ h) k ∈ N₂, since N₂ is normal in G,
and h⁻¹ k⁻¹ hk = h⁻¹ (k⁻¹ h k) ∈ N₁, since N₁ is normal in G.
Hence, h⁻¹ k⁻¹ hk = e (since N₁ ∩ N₂ = {e}). ==> hk = kh.
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Now, define the mapping f : G → N₁ x N₂, by f(g) = (h, k), where g = hk (which is unique by above).
f is 1-1, by the fact that g = hk is a unique representation.
Since |G| = |N₁| |N₂| = |N₁ x N₂| is finite, we see that f is automatically onto as well.

To show that f is a homomorphism, let g = hk, g' = h'k' be two elements in G
(with h, h' ∈ N₁ and k, k' ∈ N₂).

So, f(gg')
= f(hk * h'k')
= f(hh' * kk'), since N₁ and N₂ commute
= (hh', kk')
= (h, k) * (h', k')
= f(hk) f(h'k')
= f(g) f(g'), as required.

Hence, f is an isomorphism, and we are done.
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I hope this helps!