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I need to show that if $H$ and $K$ are subgroups of $G$, then $H \cap K$ is also a subgroup.

My working:

$I \in H, I \in K$, $I$ is unique $\implies I \in H \cap K$, so identity present in $H \cap K$

$h_c \in H \implies h_c^{-1}\in H$ and

$h_c \in K \implies h_c^{-1}\in K$ so

$h_c \in H, h_c \in K \implies h_c \in H \cap K$ and

$h_c \in H, h_c \in K \implies h_c^{-1} \in H \cap K$

so inverse of every element present in $H \cap K$.

$h_c, k_c \in H \implies h_ck_c \in H$

$h_c, k_c \in K \implies h_ck_c \in K$

$h_c, k_c \in H, h_c, k_c \in K \implies h_c, k_c \in H \cap K$

$h_c k_c \in H, h_c k_c \in K \implies h_c k_c \in H \cap K$

so $H \cap K$ is closed.

Is this correct? Do I need to say anything else?

zabop
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    This is definitely a dupe. Have you tried searching a bit before asking? – mathreadler Sep 21 '18 at 19:33
  • (I didnt find it written it like this, but you are right in suggesting that I might not invested enough effort in researching.) – zabop Sep 21 '18 at 19:37
  • Its probably only been written like this 30 times over the last 3 years or so.. Same result every time so far. – mathreadler Sep 21 '18 at 19:39

1 Answers1

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This is good, no need to say more! You have a typo in the fifth line of your reasoning (replace the first $h_c^{-1}$ by $h_c$). Also, as a matter of style, I recommend not using the subscript $c$ at all.

You could also prove that $H\cap K$ satisfies the subgroup criterion where you show that $H\cap K \ne \varnothing$ and if $h,k\in H\cap K$, then $hk^{-1} \in H\cap K$.

Alex Ortiz
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