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## Homework Statement

[PLAIN]http://img689.imageshack.us/img689/3047/directproduct.png [Broken]

[itex]<[/itex] denotes a subgroup.

[itex]\triangleleft[/itex] denotes a normal subgroup.

## The Attempt at a Solution

Have I done (a) correctly?

[itex]0 \in A[/itex] so [itex]A \neq \emptyset[/itex]

If [itex]a=x+ix[/itex] and [itex]b=y+iy[/itex]

then [itex]ab^{-1} = x-y + ix - iy = x-y + i(x-y) \in A[/itex]

[itex]\therefore A < G[/itex]

Again [itex]0 \in B[/itex] so [itex]B \neq \emptyset[/itex]

If [itex]c=x+ix[/itex] and [itex]d=y+iy[/itex]

then [itex]cd^{-1} = x-y - ix - iy = x-y - i(x+y) \in B[/itex]

[itex]\therefore B < G[/itex]

If [itex]g=x+iy \in G[/itex] and [itex]a=c+ic \in A[/itex]

then [itex]gag^{-1} = x+iy + c+ic -x-iy = c+ic \in A[/itex]

[itex]\therefore A \triangleleft G[/itex]

If [itex]b=c-ic \in B[/itex]

then [itex]gbg^{-1} = x+iy + c-ic -x-iy = c-ic \in B[/itex]

[itex]\therefore B \triangleleft G[/itex]

It is clear that [itex]A \cap B = \{0\}[/itex] .

If [itex]a=c+ic \in A[/itex] and [itex]b=d-id \in B[/itex]

than [itex]ab = c+ic +d-id = c+d + i(c-d)[/itex]

so if [itex]g=x+iy \in G[/itex] then [itex]g=ab[/itex] with [itex]x=c+d \in \mathbb{R}[/itex] and [itex]y=c-d \in \mathbb{R}[/itex]

[itex]\therefore G=AB[/itex] and G is the internal direct product of A and B.

Can anyone help with (b)?

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